mm-3619 === Subject: orthogonal matrix+decomposition Is it true that every orthgogonal matrix over R(C) we can repsresent as a product of at most two involutions? === Subject: Re: orthogonal matrix+decomposition at least it is true if det(O)=1. Think that one can find a basis where the orthogonal transformation has a matrix with only 2x2 rotation matrices on the diagonal and 1's. (because det(O)=1, one can group the -1 by pairs). And because a rotation is the product of 2 symetries, it works. Fedor. === Subject: Re: orthogonal matrix+decomposition > at least it is true if det(O)=1. Think that one can find a basis where > the orthogonal transformation has a matrix with only 2x2 rotation > matrices on the diagonal and 1's. (because det(O)=1, one can group the > -1 by pairs). And because a rotation is the product of 2 symetries, it > works. > Fedor. but could you please give an accurate definition of symmetry matrix(i've encountered with it recently and would like to have a definition) === Subject: Re: orthogonal matrix+decomposition > at least it is true if det(O)=1. Think that one can find a basis where > the orthogonal transformation has a matrix with only 2x2 rotation > matrices on the diagonal and 1's. (because det(O)=1, one can group the > -1 by pairs). And because a rotation is the product of 2 symetries, it > works. > Fedor. > but could you please give an accurate definition of symmetry > matrix(i've encountered with it recently and would like to have a > definition) let a_i_j represent the element in row i and column j of the matrix A. then A is symmetric when a_i_j = a_j_i for all i and j . === Subject: Re: orthogonal matrix+decomposition at least it is true if det(O)=1. Think that one can find a basis where > the orthogonal transformation has a matrix with only 2x2 rotation > matrices on the diagonal and 1's. (because det(O)=1, one can group the > -1 by pairs). And because a rotation is the product of 2 symetries, it > works. > > Fedor. > but could you please give an accurate definition of symmetry > matrix(i've encountered with it recently and would like to have a > definition) > let a_i_j represent the element in row i and column j of the matrix A. > then A is symmetric when a_i_j = a_j_i for all i and j . No, i think Fedor meant the other defintion of a symmetry matrix. Note, you gave the defintion of a symmetric matrix, but i wanted symmetry matrix. As Fedor used thi word insted of the word involution i think a symmetry matrix is the same as invoultion, i.e a matrix which satisfy A^2 = I, but i think that since it contains the word symmetry there is another defintion, more natural. === Subject: Re: Statistics questions <44892F59.1030707@netscape.net This is completely off topic, but do you have any idea where the name > Donjon is coming from? Check at http://www.etymonline.com/index.php === Subject: not algebraically closed field and polynomial Let K be a field, which is not algebraically closed. How can i rpove that there exists a natural number n and a homogenous polynomial f from K[x_1,x_2,...,x_n] such that the equation f(x_1,x_2,...,x_n) = 0 has only trivial(zero) solution. Thansk === Subject: Re: not algebraically closed field and polynomial > Let K be a field, which is not algebraically closed. How can i rpove that there exists a natural number n and a homogenous polynomial f from K[x_1,x_2,...,x_n] such that the equation f(x_1,x_2,...,x_n) = 0 has only trivial(zero) solution. > Thansk Arturo showed us a solution with n = 1. Check this out for a solution with n = 2. Since K is not algebraically closed, there must be an irreducible polynomial in f(x) K[x] of degree d>1. Turn this into a homogeneous polynomial in 2 variables, f(x, y), by padding it out with the right powers of y to make it homogeneous of degree d. For instance x^3 + x - 7 becomes x^3 + xy^2 - 7y^3. Now if f(x, y) = 0 and y is non-zero, then dividing the polynomial by y^d shows that f(x/y) = 0 contradicting the irreducibility of f(x). But if y = 0, then x^d = 0, so x = 0, and this is what I would call a trivial solution. Achava === Subject: Re: not algebraically closed field and polynomial > Let K be a field, which is not algebraically > closed. How can i rpove that there exists a natural > number n and a homogenous polynomial f from > K[x_1,x_2,...,x_n] such that the equation > f(x_1,x_2,...,x_n) = 0 has only trivial(zero) > solution. > Thansk > Arturo showed us a solution with n = 1. Check this > out for a solution > with n = 2. Since K is not algebraically closed, > there must be an > irreducible polynomial in f(x) K[x] of degree d>1. > Turn this into a > homogeneous polynomial in 2 variables, f(x, y), by > padding it out with > the right powers of y to make it homogeneous of > degree d. For instance > x^3 + x - 7 becomes x^3 + xy^2 - 7y^3. Now if f(x, > y) = 0 and y is > non-zero, then dividing the polynomial by y^d shows > that f(x/y) = 0 > contradicting the irreducibility of f(x). But if y = > 0, then x^d = 0, > so x = 0, and this is what I would call a trivial > solution. > Achava === Subject: Re: not algebraically closed field and polynomial days. My association with the Department is that of an alumnus. >Let K be a field, which is not algebraically closed. How can i rpove >that there exists a natural number n and a homogenous polynomial f >from K[x_1,x_2,...,x_n] such that the equation f(x_1,x_2,...,x_n) = 0 >has only trivial(zero) solution. Set n=1, f(x)=x. I suspect, however, that what you asked was not what you meant to ask. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: not algebraically closed field and polynomial > <17722098.1149878130864.JavaMail.jakarta@nitrogen.math > forum.org>, >Let K be a field, which is not algebraically closed. > How can i rpove >that there exists a natural number n and a > homogenous polynomial f >from K[x_1,x_2,...,x_n] such that the equation > f(x_1,x_2,...,x_n) = 0 >has only trivial(zero) solution. > Set n=1, f(x)=x. > I suspect, however, that what you asked was not what > you meant to ask. Really, you are right, i meant to ask another problem: Here is the right version How can i prove that there exists a natural number n and a homogenous polynomial f of degree n > 1 from K[x_1,x_2,...,x_n] such that the equation f(x_1,x_2,...,x_n) = 0 has only trivial(zero) solution. === Subject: Re: not algebraically closed field and polynomial days. My association with the Department is that of an alumnus. I don't have an answer for you, but I do have a correction for you: >> <17722098.1149878130864.JavaMail.jakarta@nitrogen.math >> forum.org>, >>Let K be a field, which is not algebraically closed. >> How can i rpove >>that there exists a natural number n and a >> homogenous polynomial f >>from K[x_1,x_2,...,x_n] such that the equation >> f(x_1,x_2,...,x_n) = 0 >>has only trivial(zero) solution. >> Set n=1, f(x)=x. >> I suspect, however, that what you asked was not what >> you meant to ask. >Really, you are right, i meant to ask another problem: >Here is the right version >How can i prove that there exists a natural number n and a homogenous >polynomial f of degree n > 1 from K[x_1,x_2,...,x_n] such >that the equation f(x_1,x_2,...,x_n) = 0 has only trivial(zero) >solution. Judging from your other response, this is also ->NOT<- what you meant to ask. You told the Lovable Snake that you wanted the number of variables to be equal to the degree of f; your statement here only requires the number of variables to be AT MOST the degree of f. So you also must add the condition that the degree of f in each of x1,....,xn must be at least 1 (i.e., that each of the variables ->actually<- occur in f). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: not algebraically closed field and polynomial > <3464942.1149925213333.JavaMail.jakarta@nitrogen.mathf > orum.org>, > I don't have an answer for you, but I do have a > correction for you: > <17722098.1149878130864.JavaMail.jakarta@nitrogen.math >> forum.org>, >>Let K be a field, which is not algebraically > closed. >> How can i rpove >>that there exists a natural number n and a >> homogenous polynomial f >>from K[x_1,x_2,...,x_n] such that the equation >> f(x_1,x_2,...,x_n) = 0 >>has only trivial(zero) solution. >> Set n=1, f(x)=x. >> I suspect, however, that what you asked was not > what >> you meant to ask. >Really, you are right, i meant to ask another > problem: >Here is the right version >How can i prove that there exists a natural number n > and a homogenous >polynomial f of degree n > 1 from > K[x_1,x_2,...,x_n] such >that the equation f(x_1,x_2,...,x_n) = 0 has only > trivial(zero) >solution. > Judging from your other response, this is also > ->NOT<- what you meant > to ask. You told the Lovable Snake that you wanted > the number of > variables to be equal to the degree of f; your > statement here only > requires the number of variables to be AT MOST the > degree of f. > So you also must add the condition that the degree of > f in each of > x1,....,xn must be at least 1 (i.e., that each of the > variables > ->actually<- occur in f). Yes, again i'm sorry i wanted: How can i prove that there exists a natural number n and a homogenous polynomial f of degree n > 1 in n variables from K[x_1,x_2,...,x_n] such that the equation f(x_1,x_2,...,x_n) = 0 has only trivial(zero) solution. === Subject: Re: not algebraically closed field and polynomial >> <3464942.1149925213333.JavaMail.jakarta@nitrogen.mathf >> orum.org>, >> I don't have an answer for you, but I do have a >> correction for you: > <17722098.1149878130864.JavaMail.jakarta@nitrogen.math > forum.org>, >Let K be a field, which is not algebraically >> closed. > How can i rpove >that there exists a natural number n and a > homogenous polynomial f >from K[x_1,x_2,...,x_n] such that the equation > f(x_1,x_2,...,x_n) = 0 >has only trivial(zero) solution. >> Set n=1, f(x)=x. >> I suspect, however, that what you asked was not >> what > you meant to ask. >>Really, you are right, i meant to ask another >> problem: >>Here is the right version >>How can i prove that there exists a natural number n >> and a homogenous >>polynomial f of degree n > 1 from >> K[x_1,x_2,...,x_n] such >>that the equation f(x_1,x_2,...,x_n) = 0 has only >> trivial(zero) >>solution. >> Judging from your other response, this is also >> ->NOT<- what you meant >> to ask. You told the Lovable Snake that you wanted >> the number of >> variables to be equal to the degree of f; your >> statement here only >> requires the number of variables to be AT MOST the >> degree of f. >> So you also must add the condition that the degree of >> f in each of >> x1,....,xn must be at least 1 (i.e., that each of the >> variables >> ->actually<- occur in f). >Yes, again i'm sorry i wanted: >How can i prove that there exists a natural number n and a homogenous polynomial f of degree n > 1 in n variables from K[x_1,x_2,...,x_n] such that the equation >f(x_1,x_2,...,x_n) = 0 has only trivial(zero) >solution. Over a field lacking sqrt(7), consider x1^2 - 7 x2^2. Over a field lacking cbrt(2), consider x1^3 + 2x2^3 + 4x3^3 - 6 x1 x2 x3. Figure out the pattern and generalize it. -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: optimization <128i31hk8ml4k64@corp.supernews.com> <128irhmcas2f8c5@corp.supernews.com> <128itp1ru7st9bc@corp.supernews.com> <128jclaeaj3fk8e@corp.supernews.com> let f(x; t) be a function f:R->R with parameter t in R who has no >> local >> extrema given by the derivative method. >> sup [x in [a,b]] sup[t in [c,d]] f(x; t) = sup([alpha, beta) in >> boundary >> of [a,b]X[c,d]] f(alpha; beta) > Are you wanting local extrema for function of one or two variables? > Yes, f(x,t) = x + t has no extrema unless you restrict f(x,t) to a > bounded region of it's domain. Are you wanting the extrema of > x + t, or the extremas of x + a for each value of a in domain t? > Now the parameter idea, I suppose, can simply be looked at as an >> independent variable. >> Lets suppose I have the function >> f(x,y) = 1/3*x^3 - 1/2*y*x^2 + 1/4*x >> and I want to find the global maximum. >> f_x = x^2 - yx + 1/4 >> f_y = -x^2/2 >> f_x = f_y = 0 gives >> {x = 0, y in R} >> but if the domain is the square [4,9]x[0,1] this extrema is not in >> it >> so >> the >> only other options are the points > Look out for saddles. > You have to check if along all four of those lines there are no > extrema. > f_x(x,0) > f_x(x,1) > f_t(4,t) > f_t(9,t) > (4,0), (4,1), (9, 0), (9,1) >> f(4,0) = 67/3 >> f(4,1) = 43/3 >> f(9,0) = 981/4 >> f(9,1) = 819/4 >> So the maximum of f(x,y) in the square is 981/4. >> My original point had to do with the fact that I know that there >> are >> no >> critical points in my domain so they must all occur on the >> boundary. >> What boundary? What if boundary = nulset as if domain is R or R^2? >> Obviously then I am talking about a bounded subset of R or R^2 then > then >> If I remember correctly this is basic calculus where you would have >> to >> find >> extrema but if they didn't occur then they would occur at the >> intervals. >> I >> just don't remember any formulation of it except by intuition. I'm >> looking >> for a more formula approach. > Nonsense, some functions have no extrema. However continuous > functions > over a compact domain will attain their max and min. >> huh? Every bounded function must have an extrema on a bounded set... > it > might not be unique but there is one there. >> if f:R^n->R and grad(f) != 0 in some domain then the global max >> occurs >> on >> the boundary of that domain. I suppose one has to consider saddle >> points >> and >> the like though ;/ > bd R^n = nulset. f(x,y) = x + y defined over R^2 has no extrema > and the boundary of the domain, bd R^2, is empty. >> Obviously I'm talking about a bounded subset of R or R^2. >> I also probably, most likely since that is all I have talked about, > would > be > talking about a closed subset of R... i.e. a compact set. >> You keep refering to R.. sorry for not being clear but just because I > wasn't > you don't have to be anal about it. > f(x,y) = x + y with domain Rx[0,1] and bd Bx[0,1] = Rx{ 0,1 } > has no extrema because the domain isn't compact or bounded. > yeah... so why you keep talking about it then? > f(x) = 1/x defined over R0 has no extrema. > R0 isn't compact even tho bd R0 = {0} > and f'(x) /= 0 for all x in domain. >> So your nonsense about intervals doesn't compute. >> Um... you keep refering to looking at all of R.. > jesus christ. Did I ever mention looking for an extrema over R? All my > examples were using compact sets... chances are that is what I'm > talking > about, no? >> Sheesh.. are you always this anal? Next time I make a mistake if you > are > going to jump all over me and not assume the obvious then please don't > reply. This isn't a math test and while I might have made mistakes or > not > have been clear at the very least you could assume what most likely I > was > talking about. >> and to make your points irrelevant: >> if f:R^n->R then max(f(x), x in R) := sup(f(x), x in R). > (notice the :=) >> so if I want, I can just talk about oo being the maximum if that is > the > case... makes no big deal to me... technically the maximum and sup are > different but in my case they are not... >> I'm sure all the talk about boundries and such would imply that the > domain > I'm interested actually has a boundary, No? Unless you believe that I > believe R has a non-empty boundary? Now I suppose you will bitch that > I > am > not saying it is closed too? >> Given various misstatements in your original post, I wouldn't >> press too hard on the obligation of responders to make sense >> your question(s). >> Theres a difference between making sense and beeing an asshole about it. >> Short answer is, yes, an extremum of a function that is >> continuous on a closed domain and differentiable on the >> interior of the domain must either occur at a point in the >> interior where the derivative (or gradient, if you want more >> than one dimension) or at a point on the boundary. >> differentiability has nothing to do with it. The function doesn't even >> have >> to be continuous... and bounded only if you feel that the extrema must >> occur >> on a bounded domain. >> There are functions that are not differentiable that take on a max(a true >> max). >> I can construct one quite simple >> f(x) = >> 10 if x = 0 >> 1 if x is rational >> 0 if x is irrational >> obviously f is not differentiable or even continuous. >> but surely max(f(x)) on R is 10? >> what about >> g(x) = >> exp(-|x|) if x is rational >> 0 if x is irrational >> max(g(x))? >> ok, what about >> h(x) = >> 0 if x = 0 >> exp(-|x|) if x is rational >> 0 if x is irrational >> max(h(x))? (it might not exist but sup(h(x)) = max(g(x)) >> So I am making these off the top of my head... maybe I made a mistake >> somewhere... so sue me. >> (note that sometimes though I think of max as the same as sup... which >> could >> cause some problems but tough...) >> More generally you can ask just that the function be >> continuous on a closed domain. If an extremum occurs >> in the interior of the domain, the derivative or gradient >> will either be zero there or undefined. This is the >> definition of critical point. For example, the minimum >> of the absolute value function on [-10,10] occurs at a >> critical point (derivative undefined at zero), while the >> maximums are in this particular case the endpoints. >> Yes, and the maximum of the step function on [2, 5] is at each point in >> [2,5]... none of which the the derivative will tell you... infact the >> derivative is pretty much useless for finding extrema on compact subsets >> for >> this function. Ofcourse this is an easy function. >> My question, which I stated many times in enough of a way that one could >> understand what I'm talking about if they put forth any effort, even if >> it >> wasn't perfect or even close was simply that: >> if all critical points of a function are non-real then any extrema must >> occur on the boundary. >> This has nothing to do with R like William wants me to believe. If the >> domain is R then obviously the boundary set is empty and hence there are >> no >> critical points and neither any extrema. The statement above, if taken >> as a >> true implication is perfectly valid if you are dealing with the domain >> being >> R.... it just happens to be vacuously true then. >> There is no reason for William Elliot to act like some buffoon just >> because >> I didn't do a good job in explaining what I ment. Its very simple to >> write >> a reply such as >> This makes no sense to me. Can you please rethink your question. >> or something like that. >> Instead he continuously focuses on the fact that in R one has no boundary >> and hence f cannot attain its maximum on it. >> Infact R does have a boundary and it is the empty set and hence there is >> no >> way f can obtain its maximum on it sense there is no elements in it. >> So whats his point? He thought he was being smart by trying to make me >> look >> stupid(whcih I can do that by myself). >> An appropriate argument would have been that f cannot obtain its max on a >> non-closed set. This takes care of the boundary problem on R and on >> other >> sets. Opps, I made a mistake... I didn't say that f had no global max on >> its >> boundary. Sorry... don't chew me out. >> You know, I'm not taking a test or writing a calculus book so there is no >> need to try and be perfect. I'm asking a simple question that you >> obviously >> got.. if my errors were so bad that it was unintelligable then just say >> so. >> I would have no problem trying to be more precise and clear. But if you >> want to basicaly bitch me out for it then don't even bother. I'll figure >> it >> out on my own. > Believe it or not, in trying to point out gaps or simple mistakes, > I'm not chewing you out. I don't know you and don't presume > to have walked a mile in your shoes, so let's not go there. > No, not you but William but you seemed to say his behavior is acceptable. > You atleast tried to be helpful. > I do know the definition of critical point, and I suspect you are > getting off-track in thinking about if critical points of a function > are non-real. First of all, since we want to discuss extrema, > we should probably restrict discussion to real-valued functions. > Complex numbers are not an order field, so we really can't say > much about extrema for complex-valued functions unless we > segue to their real-valued modulus. Second of all if we want to > talk about real-valued functions over a complex domain, what > we really need to do is think about the domain as if it were a > multi-dimensional real domain. > Again, critical points are places where the derivative (or gradient > in a multi-dimensional domain) is zero OR undefined. Of course > if you pick a function, like your step function example, which has > a lot of critical points, then the theorem that tells us extrema > occur at critical points or on the boundary is less helpful than > in case there are only a few critical points to consider. > If there are no critical points and the boundary is empty, then > there are no extrema. So, even if you thought William was > belaboring the obvious, there is something there to think on. > I'm all in favor of your figuring it out on your own. At best we > can help by introducing you to some clear reasoning or spot > some rough patches in your presentation, but even then it's > your achievement to sort things out and make it your own. > Three stages of learning something in math: > 1. Understanding someone's explanation. > 2. Applying it for yourself to a new problem. > 3. Explaining it to someone else's satisfaction. > best wishes, chip > lol. Yeah, I should know all that stuff to as I spend about 10 years > studying math and have a BS in it. Ofcourse that was many years ago and > I've forgotten alot. > My point was simply this: > I had function I was trying to optimize. Ok, what does that mean? Basicaly > find the extrema(the maximum value actually). How is that done? Find the > critical points? Ok, so I looked for critical points and what did I find? I > found that there were none. But how did I know? Well just refer to the first > example in my second post. Now the only other choice is to look on the > boundary? > Its very simple: The extrema are on the boundary and/or in the interior. I > have excluded them on the interior by show there are no critical points in > the domain of definition... hence they must occur on the boundary? I was > just trying to clarification of this incase I forgot something. > Sure I didn't mention some specifics that may or may not be important such > as the domain being compact... I didn't mention it because I forgot... > although my examples should have probably hinted at that since they were all > compact. > But the thing that confused William and that he went off ranting about R > having no boundary which doesn't matter was that I was dealing with a > function that had a parameter. > i.e. > f(x; t) > I was able to show that in my domain that there are no critical points for > f(x; t) for all t in the parameters domain. > Hence I just have to look for extrema on the boundary of the domain. > I will give the example a second time > f(x; t) = 1/3*x^3 - 1/2*t*x^2 + 1/4*x > now, if you were asked to find the extrema of that what would you do? The > obvious thing would be to find the critical points, right? (even without > knowing the domain of x and t yet that tends to be the first thing). > i.e., > f'(x; t) = x^2 - t*x + 1/4 = 0 > == x = (t +- sqrt(t^2 - 1))/2 > but if I know said |t| < 1 then you automatically know that there are no > critical points because x is not real. so, if you want to think about it > like this > the family of real valued functions {f(x; t} | |t| < 1} have no critical > points. > Very simple... and I didn't even have to mention the domain of x, it is true > for all subsets of R. > So the next obvious thing is to look at the boundary of the domain of x. If > the domain of x is R then the boundary is empty so there are no extrema. > (And which William thinks that it really matters). If the boundary of the > domain of x is not empty then one must search over it to find the extrema of > of the family. If the boundary is not closed then there might not be a max > but the sup will exist as long as f(x; t) is bounded. > For my problem, which I didn't state the first time, I am strictly dealing > with compact sets. It doesn't matter all that much though as the process > tends to be the same. While you could treat t as an independent variable > for my purposes I'll treat it as a parameter. Why? Because the problem I'm > dealing with is more complicated and I am making a simplification. In this > case it works either way but makes more sense in my original problem to > treat it as a parameter. > Me: > My original point had to do with the fact that I know that there are no > critical points in my domain so they[the extrema] must all occur on the > boundary. > William: > What boundary? What if boundary = nulset as if domain is R or R^2? > so whats his point? If the boundary is empty then obviously there cannot be > any extrema there but it doesn't change anything... its vacuously true that > there are no extrema on an empty set. > Basicaly he's saying that R has no boundary because the boundary of R is the > empty set. This isn't even a pedantic interpretation of boundary... Theres > no reason to say that R doesn't have a boundary just cause it happens to be > the empty set. Actually both ideas are equivilent but it doesn't change > anything I said. That is, it doesn't prove anything I have said was > wrong(Even though all my mistakes and such were more than enough). > Now if he said if the domain was not closed that there could be no maximum > on the boundary then that would have been a different story. i.e., I use max > several times but its not necessarily true that there is a max for a general > subset of R(but theres always a sup). > Anyways, This isn't going anywhere... Don't think I am jumping on you > though. That is not my intention. I'm just trying to point out that William > is trying to put up this huge argument as a counter example of what I was > saying when it is fallacious. Sure there are lots of holes in what I said > and lots of mistakes and typos. > For example, > William says > f(x,y) = x + y with domain Rx[0,1] and bd Bx[0,1] = Rx{ 0,1 } > has no extrema because the domain isn't compact or bounded. > Should I point out the error that compact or bounded is the same as > compact? > He is correct in what he says though as by extrema one tends to mean that > the function must exist at that point. > In my case I don't care if it exists there or not... I am trying to optimize > a function and so if I know the sup then it will do just fine because I can > approximate the extrema. > i.e. > f(x) = x for x in [0,4) > Theres no extrema but surely x = 4 is an extrema in some sense.. > i.e. > sup[x in [0,)](f(x)) = 4 > since I'm optimizing and I only care about good results.. > 3.9, 3.999, 3.999999 would all be good answers for me. Atleast then I have > some numbers to do some other things with. > So really what I'm after is the extrema on the closure of the domain. And > here william will bitch that there will be no extrema on the closure of R if > there are no critical points. > But what can I say? I don't know what to call extrema that are not > realizable. I want the extrema because I can approximate them and it > would be good enough for my application. f(x) = x has no critical points on > R and no extrema either... but sup(f(x)) = R... hence I can approximate it > by any arbitrarily large number. For my application maybe 1 will do, maybe > 100000. Who knows. > I never learned a word for such points... hence I just use the word extrema. The least upper bound of a set is called a supremum. The greatest lower bound, an infimum. > Should I point out the error that compact or bounded is the same as > compact? In finite dimensional Euclidean space: compact <==> closed and bounded In any case, it is certainly as important to spell out the domain as the function, when the problem is that of optimizing a function over a domain. I can tell from your function: f(x; t) = 1/3*x^3 - 1/2*t*x^2 + 1/4*x that we are dealing with two variable/dimensions (at least), but if you ask me to find the extrema, I will ask on what domain. I understood William's example to illustrate why the domain matters, but I'm sure that the importance of the domain is something we all agree on. It can be a pain to search on the boundary for extrema. It's also a pain to sort out the absolute extrema from the local ones, unless they are few in number and one can simply pick the biggest/smallest values. But that's why us math workers get paid the big bucks, right?? === Subject: Re: optimization So really what I'm after is the extrema on the closure of the domain. And >> here william will bitch that there will be no extrema on the closure of R >> if >> there are no critical points. >> But what can I say? I don't know what to call extrema that are not >> realizable. I want the extrema because I can approximate them and it >> would be good enough for my application. f(x) = x has no critical points >> on >> R and no extrema either... but sup(f(x)) = R... hence I can approximate >> it >> by any arbitrarily large number. For my application maybe 1 will do, >> maybe >> 100000. Who knows. >> I never learned a word for such points... hence I just use the word >> extrema. > The least upper bound of a set is called a supremum. yeah, but extrema includes all local extrema... there is only one supremum ;/ so I can't talk about the set of lead upper bounds. So one has to group the supremum and the infimum along with the critical points to get the set I'm talking about. > The greatest lower bound, an infimum. >> Should I point out the error that compact or bounded is the same as >> compact? > In finite dimensional Euclidean space: > compact <==> closed and bounded > In any case, it is certainly as important to spell out the domain > as the function, when the problem is that of optimizing a function > over a domain. yep, which is what we are talking about. Sure. I agree its important... but just because I don't state it explicity or forget to doesn't mean I deserve to be jumped on. Its much nicer to ask What is the domain? > I can tell from your function: > f(x; t) = 1/3*x^3 - 1/2*t*x^2 + 1/4*x > that we are dealing with two variable/dimensions (at least), but > if you ask me to find the extrema, I will ask on what domain. Well, lets say you were investigating this function for the first time on your own... and there no one gave you a domain. Would you just quite and give up? Maybe go around asking someone if they could give you a domain? I'd expect you would try the most obvious domain you could think of, right? Surely we are not talking about the domain of mentally challenge people X matricies who's 3rd column has a 1 in it. Its natural to look at the domain that makes the most sense. In the above example one would probably choose R or an arbitrary subset of R. (obviously finding the extrema on R will help in finding the extrema on any bounded subset of R). For example, given that function above I can easily find all the extrema on any subset of R... doesn't matter if its bounded or not. So its not that big of a deal. Not that its not an issue... but again, this isn't a math test. > I understood William's example to illustrate why the domain > matters, but I'm sure that the importance of the domain is > something we all agree on. Sure. But as I stated above, it doesn't matter as if one is talking about a bounded subset of R or R itself... or even if the set is compact. The *Method* is whats important and is applicable in all cases. In fact the method is what I was asking about and not about domains. (because obviously atmost I am dealing with R^n). Now maybe I used max when I should have used sup and called a least upper bound that didn't exist in the domain an extrema but thats just to bad. I'm allowed mistakes arent I? Not that mistakes are good and they should be pointed out but being an ass is uncalled for. > It can be a pain to search on the boundary for extrema. Sure. > It's also a pain to sort out the absolute extrema from > the local ones, unless they are few in number and > one can simply pick the biggest/smallest values. For my problem I am doing it numerically so its easy either way. Unfortunately my problem is that it can take an awful long time. So I did some math and found the critical points. Since I had a parameter and that parameter showed up in the equation for the critical points and forced there to be no solution I know that the domain my parameter gives no critical points for the function. Hence I just have to search on the boundary instead of the interior. Its much faster this way... the hard part is to describe the boundary in an efficient way. > But that's why us math workers get paid the big bucks, > right?? I don't know... maybe. Jon === Subject: Re: optimization <128i31hk8ml4k64@corp.supernews.com> <128jkmknh9d8r40@corp.supernews.com f(x; t) = 1/3*x^3 - 1/2*t*x^2 + 1/4*x > Well, lets say you were investigating this function for the first time on > your own... and there no one gave you a domain. Would you just quite and > give up? Maybe go around asking someone if they could give you a domain? Nope, we'd assume the naturally implied domain which is this case is R^2, just as the naturally implied domain of f(x) = 1/x is R0. > I'd expect you would try the most obvious domain you could think of, right? > Surely we are not talking about the domain of mentally challenge people X > matrices who's 3rd column has a 1 in it. A thin skinned cusser who complains about being jumped on, must be mentally challenged to overlook how his causal, nay impolite use of language, is asking for it. > Its natural to look at the domain that makes the most sense. In the above > example one would probably choose R or an arbitrary subset of R. (obviously > finding the extrema on R will help in finding the extrema on any bounded > subset of R). > For example, given that function above I can easily find all the extrema on > any subset of R... doesn't matter if its bounded or not. The given function with it's implicit domain has no global extrema. > So its not that big of a deal. Good, then we can drop it. === Subject: Re: stat question >>Could someone help me with this. I have two normal distributions with >>mean x1 and mean x2 with standard deviations of s1 and s2. What is the >>standard deviation of the ratio x1/x2? >>This is a real problem. If x1 < x2 and x1 and x2 are normally >>distributed, it seems clear that the ratio would NOT be normally >>distributed. What would we call the distribution? Plotting it out it >>looks log normal but I can't prove this. >> If your assumptions are correct, then x1 and x2 are not >> independent, so it is difficult to give an answer. In any >> case, it would be difficult, and I do not even believe >> possible, for the ratio to have a mean, let alone a >> variance. It cannot be lognormal. >OK I can accept this last fact. Let me restate my original question >about the ratios with an example. If I can get my head around this >example the real problem can be tackled. >Let x1 be the number of disintegrations in 5 min for some radioactive >isotope. Not normal, Poisson. > s1 will be the sq rt of known dpm times 5 min. Let x2 be the >number of disintegrations in 20 min so s2 wil be sq rt of known dpm >times 20 min. Does the 20 minute interval include the 5 minute interval? If not, why do you think x1 < x2? > I repeat this measure N times sometimes for shorter or >longer periods of time but that time for x1 is always 1/4 the time for >x2. For each pair the ratio of x1/x2 is computed. The ratio x1/x2 is not defined if x2 = 0. Since the probability of this occurring is nonzero (though perhaps small), x1/x2 does not have a mean or standard deviation. > It seems obvious >that the average of ratios will approach exactly 1/4. Why can't there >be a std dev associated with that ratio that is related to s1 and s2? Suppose X1 and X2 are independent Poisson random variables (which they would be if the time intervals were disjoint) with means r and 4 r respectively. To avoid the problem of division by 0, consider X1/(1+X2) rather than X1/X2. Then, according to Maple, E[X1/(1+X2)] = 1/4 - exp(-4 r)/4 and Var[X1/(1+X2)] = -1/4 (r + 1) exp(-4 r) (gamma + ln(-4 r) + Ei(1, -4 r)) - 1/16 (1 - exp(-4 r))^2 = 5/(64 r) + 3/(128 r^2) + 7/(512 r^3) + 3/(256 r^4) + ... as r -> infty. Thus the standard deviation of X1/(1+X2) is sqrt(5)/(8 sqrt(r)) + 3 sqrt(5)/(160 r^(3/2)) + 61 sqrt(5)/(6400 r^(5/2)) + 1017 sqrt(5)/(128000 r^(7/2)) + ... as r -> infty. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: JSH: SF: Does this work? [Tim Peters, to JSH] > ... > I showed a different way in which you _could_ have gotten confused into > believing the result allowed two solutions for y, and you'd get a lot > more mileage out of embracing that way because the error is subtler > (indeed, you could even tie it into another confused rant about how sqrt() > has two results in reality). The mistake you actually made was easily > explained. A cleverer mistake is just waiting for you to find it. While it's probably futile, I wanted to point out that this other way has nothing to do with completing the square: it follows directly from your original equations (which won't be a surprise to anyone else, since completing the square doesn't _change_ anything, it just rearranges the symbols). This is pure cut-and-paste of a complete Macsyma session, so there are no transcription errors or hidden steps. c lines are input, d lines are output, % in an input line refers to the result on the preceding d line: (c1) fancy_display: false (d1) false (c2) k = -21*z^2 - 10*y*z + 3*f_2*z - 19*f_1*z - y^2 + f_2*y - 5*f_1*y + 2*f_1*f_2 - 4*(f_1)^2 2 2 (d2) k = - 21 z - 10 y z + 3 f_2 z - 19 f_1 z - y + f_2 y - 5 f_1 y 2 + 2 f_1 f_2 - 4 f_1 (c3) %-rhs(%) 2 2 (d3) 21 z + 10 y z - 3 f_2 z + 19 f_1 z + y - f_2 y + 5 f_1 y + k 2 - 2 f_1 f_2 + 4 f_1 = 0 (c4) subst(k = f_1*f_2 - g_1*g_2, %) 2 2 (d4) 21 z + 10 y z - 3 f_2 z + 19 f_1 z + y - f_2 y + 5 f_1 y 2 - g_1 g_2 - f_1 f_2 + 4 f_1 = 0 (c5) subst(z = -(3*f_1 - f_2 + g_1 - g_2)/4, %) 2 5 (- g_2 + g_1 - f_2 + 3 f_1) y (d5) y - ------------------------------- - f_2 y + 5 f_1 y - g_1 g_2 2 2 21 (- g_2 + g_1 - f_2 + 3 f_1) 3 f_2 (- g_2 + g_1 - f_2 + 3 f_1) + ------------------------------- + --------------------------------- 16 4 19 f_1 (- g_2 + g_1 - f_2 + 3 f_1) 2 - ---------------------------------- - f_1 f_2 + 4 f_1 = 0 4 (c6) %,factor (d6) (4 y + 3 g_2 - 7 g_1 + 3 f_2 - 5 f_1) (4 y + 7 g_2 - 3 g_1 + 3 f_2 - 5 f_1)/16 = 0 (c1) tells the system to use ASCII art for display instead of nice mathematical typesetting. (c2) is your equation for k, after expanding it. (c3) just moves everything to the LHS. (c4) replaces k with your equation for it in unexpanded form. (c5) replaces z with the solution for z obtained from your original 4 equations. (c6) just factors the result, and you can see from the product that you do indeed get exactly the two solutions for y I said you'd get if you did a similar thing starting from the complete-the-square respellings Rick and I derived for you. While it's not a lot of fun, you can easily _check_ that all the steps above are in fact correct. So how do you end up with two solutions for y here? That's an interesting confusion :-) If it attracts your interest, I suggest reproducing the phenomenon with a much smaller system of equations: less tedium makes more room for insight. === Subject: Re: Correlation >I am trying to work out Correlation(X,Z) using Correlation(X,Y) and >Correlation(Y,Z). I am pretty sure that I don't have enough To see that you don't have enough information, consider the case where Y is independent of both X and Z, but X and Z are related in some way. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Correlation All you can say is that rxz must be in the interval rxy*ryz +- sqrt[(1-rxy^2)(1-ryz^2)]. === Subject: Re: Is it identity? This relates to a recent research project I am doing on paradoxes. I believe that what you are describing is an automorphism. It is a mathematical object that maps onto itself. It is a function that is not exactly a bijective mapping. A bijective mapping is something like xRy. An automorphism is more in terms of A=A. Insofar as a continuous function is concerned, there should be no fixed points. There should only be approximate vectors. Vectors describe direction and magnitude. They do not necessarily describe fixed points. Here, you might be confusing a discrete finite function with a continuum. Although, they can be transformed into each other, they are internally different mathematical structures. One of the features of a discrete function is that you can divide and add. An infinite continuum will lack this feature. This is not standard math. Now, one of the problems of an automorphism is that it falls into Curry's paradox. Curry's paradox is a type of paradox to which I call 'self-justifying.' ************************************************************************ If everything inside this box is true then tomorrow all women would stop faking orgasms. ************************************************************************ See how it self-justifies? There is no independent outside confirmation. You go outside to prove it then it is no longer inside the box. It is no longer an automorphism. It loses the identity thesis. Yes, based on the assumption that if everything in the confines of this box is true then tomorrow all women would stop faking orgasms. However, there is no way to prove this. It is unproven. In math and logic, this means that it is neither true nor false. To go outside to confirm means that it is false. You've just lost the identity thesis of the box. So how can it be **both** neither true nor false, i.e. unproven **and** proven false. There is a way out of this, but I have yet to publish the results. What an automorphism does is it exacts a circularity issue, 'if true then true,' but intuitively we know it to be false. It is a different form of circularity. It is not like 'I am lying,' where you have 'if true then false,' or 'if false then true.' I have evaluated several mathematicians who do this, and think that they have a proof. They do not have a proof. What they have a perfect monstrosity of this sort. I have written to several mathematicians on this feature. So far, none have seen the flaw. B.T. > If a continuous function from the set of real numbers to itself has fixed points almost everywhere then it is the identity function. Prove or disprove. === Subject: Re: Is it identity? > I have written to several mathematicians on this feature. > So far, none have seen the flaw. B.T. Most people would draw a different conclusion from this than you're apparently drawing. Dave L. Renfro === Subject: Re: Is it identity? Oh, by the way. Curry's paradox is a paradox about naive set theory. The example is not to put down anyone. It is more to demonstrate the vividness of the naivete of people who espouse this kind of structure as a proof: Hope no one is offended. > This relates to a recent research project I am doing on paradoxes. > I believe that what you are describing is an automorphism. It is a > mathematical object that maps onto itself. It is a function that is not > exactly a bijective mapping. > A bijective mapping is something like xRy. > An automorphism is more in terms of A=A. > Insofar as a continuous function is concerned, there should be no fixed > points. There should only be approximate vectors. Vectors describe > direction and magnitude. They do not necessarily describe fixed points. > Here, you might be confusing a discrete finite function with a > continuum. Although, they can be transformed into each other, they are > internally different mathematical structures. > One of the features of a discrete function is that you can divide and > add. > An infinite continuum will lack this feature. > This is not standard math. > Now, one of the problems of an automorphism is that it falls into > Curry's paradox. > Curry's paradox is a type of paradox to which I call > 'self-justifying.' > ************************************************************************ > If everything inside this box is true then tomorrow all women would > stop faking orgasms. > ************************************************************************ > See how it self-justifies? > There is no independent outside confirmation. You go outside to prove > it then it is no longer inside the box. It is no longer an > automorphism. It loses the identity thesis. > Yes, based on the assumption that if everything in the confines of this > box is true then tomorrow all women would stop faking orgasms. > However, there is no way to prove this. > It is unproven. In math and logic, this means that it is neither true > nor false. > To go outside to confirm means that it is false. You've just lost the > identity thesis of the box. > So how can it be **both** neither true nor false, i.e. unproven **and** > proven false. > There is a way out of this, but I have yet to publish the results. > What an automorphism does is it exacts a circularity issue, 'if true > then true,' but intuitively we know it to be false. > It is a different form of circularity. It is not like 'I am lying,' > where you have 'if true then false,' or 'if false then true.' > I have evaluated several mathematicians who do this, and think that > they have a proof. They do not have a proof. What they have a perfect > monstrosity of this sort. > I have written to several mathematicians on this feature. So far, none > have seen the flaw. B.T. > If a continuous function from the set of real numbers to itself has fixed points almost everywhere then it is the identity function. Prove or disprove. === Subject: Point group symmetry of a labelled graph from automorphism group? Anyone have any pointers on how to do that? I played with the idea in my head, and it doesn't seem that simple.... === Subject: Re: Field of fractions (was: Re: Proofs by induction and infinity) basically the right idea, that induction doesn't work up to infinity. I had realised that I could apply a similar proof to rational numbers for commutativity of addition, but was looking to have my expectation that proof by induction has the limits I thought it did either confirmed or disconfirmed. In fact I think I should look further into number theory. One problem I have in trying to fill in the holes in my maths knowledge is that most of the books (etc.) that I look at seem to assume that you can prove theorems already, and I find it frustrating when I get stuck on examples. I have proved things before while at school and at university level, for example I did do one masters level module in design and analysis of algorithms where I certainly remember quite happily proving the worst case time complexity of algorithms under exam conditions and getting a good grade at the end. And the only pure mathematics module I studied at university level [note: not studying more maths then is something I've eternally regretted] was a masters level module on resolution theorem proving, and I did well on that too. But I seem to have forgotten how to prove things not having done it for a very long time. And so even when I understand the material if I read up on, say, probability theory, it's very frustrating to find throwaway comments that some easy proof is left as an exercise for the reader, and getting stuck. Ross-c === Subject: Re: Does this kind of function exist? > This example does not work perfectly (it is not a surjection). >> Of course not. Why should it be? I never claimed that it was, and the >> other poster did not require that. > y = f(x1, x2, ... xn), x1, x2, ..., xn, y are real; satisfying the > one-to-one mapping, namely, the different vector [x1, x2, ... xn] maps > to different y-values, and vice versa. > ^^^^^^^^^^^^^^ sentence was an explanation of what that means. But I am not sure that and vice versa means that the function is surjective. Perhaps that the OP will explain what he meant, although I doubt that that will happen. Jose Carlos Santos === Subject: Re: Does this kind of function exist? Michel Hack nous a r.8ecemment amicalement signifi.8e : >> One example (bijection from [0,1[^2 in [0,1[ : > Nice -- you provided more than was asked for (an injection -- you gave > a bijection). > In fact, it is possible to do even better: a bijection from the > closed unit hypercube to the closed unit interval (yours was > half-open) Yes, my bijection is from [0,1[^2 in [0,1[. It may be extended to [0,1]^2 in [0,1] thru another bijection from [0,1[ to [0,1] f(0) = 1 f(x) = x + 1 - 3(2^Ent(log_2(x))) Explanation : You write x in binary and then toggle 0 in 1 and 1 in 0 for every digit right to the dot up to (and including) the first 1. The remainder of digits remain unchanged. > that is strictly monotonic in each dimension. (I provided > the basic bi-monotonic bijection between the closed unit square and > the closed unit interval in previous posts; higher dimensions can be > built up from that.) Surely, mine is less pretty (not monotonic) > Some would like to go even further, and ask for a continuous bijection > -- but that is impossible for topological reasons. (Continuous > surjections exist, but not injections.) Yes again, it's a pity. -- Patrick === Subject: Calculus problem I need to do this assignment from my Calc class but I have no I idea what to do with it. Here is the exact text from my worksheet: Two tractors are plowing through a cornfield. The field is in a shape of a square with an area of 3600 square feet. The first tractor, Big Birtha, enters from the left side of the field at 60 feet. The second tractor, Tiny Tonka, enters from the lower edge at 60 feet. After 1 minute, the tractors have moved to a new location. Big Birtha has gone 4 ft South and 2 feet East from its old position. Tiny Tonka has gone 2 feet north and 4 feet West from its old position. Question: Will the two tractors collide if they maintain their speeds and directions? If so, when? If not, how close do they actually come to each other? This is designed to be drawn out on a graph so the Big Birtha is comming from the y-axis and the Tiny Tonka is comming from the x-axis. If anyone could please explain a process to solve this and any advance. NOTE: I know the speeds and size of the field are unrealistic but its just to apply calc to something slightly real-life I guess. The size if the tractors is irrelevant for this, just use an exact point. === Subject: Re: Calculus problem Yes it was homework and I would have just taken care of it myself, only the math books have already been collected for the year (highschool class), so looking up formulas or figuring out how to do it by myself with no help really wouldn't work out too well. === Subject: Re: Calculus problem > I need to do this assignment from my Calc class but I have no I idea > what to do with it. Here is the exact text from my worksheet: > Two tractors are plowing through a cornfield. The field is in a shape > of a square with an area of 3600 square feet. The first tractor, Big > Birtha, enters from the left side of the field at 60 feet. The second > tractor, Tiny Tonka, enters from the lower edge at 60 feet. It's not very clear what enters from the left side of the field at 60 feet means. Ditto enters from the lower edge at 60 feet. Is it perhaps saying that the tractors start at opposite corners of the field? If this is the exact text then it's poorly worded (unless accompanied by a diagram I guess...) > After 1 > minute, the tractors have moved to a new location. Big Birtha has gone > 4 ft South and 2 feet East from its old position. Tiny Tonka has gone 2 > feet north and 4 feet West from its old position. > Question: Will the two tractors collide if they maintain their speeds > and directions? If so, when? If not, how close do they actually come > to each other? > This is designed to be drawn out on a graph so the Big Birtha is > comming from the y-axis and the Tiny Tonka is comming from the x-axis. > If anyone could please explain a process to solve this and any > advance. Start by finding the positions of the two tractors at time t, and hence the distance between them at time t. Then use calculus to find the minimum distance and the time at which that occurs. > NOTE: I know the speeds and size of the field are unrealistic but its > just to apply calc to something slightly real-life I guess. The size > if the tractors is irrelevant for this, just use an exact point. === Subject: Re: Calculus problem Hey there, ok, I think this problem is fairly simple, here's what I understand by the question (please correct me if I'm wrong)... The square has an area of 3600 sqft, so each side is 60ft long. So, in a plan view of the field, Big Bertha enters at the top left, and Tiny Tonka enters at the bottom right. Now, modelling the plan view mentioned above above as a graph with point (0,0) at the bottom left corner of the field, we know that after t minutes, Big Bertha's x coordinate is given by 2t (since she moves 2 feet east each minute), and the corresponding y coordinate is given by 60 - 4t (since she starts at the top of the field (hence the 60) and moves 4 feet south every minute (hence the -4t)). Similarly, after t minutes, Tiny Tonka's x coordinate is given by 60 - 4t and the corresponding y co-ordinate is given by 2t. To prove the two tractors collide, we need to show that for some value of t, the coordinates of each tractor are the same. If we assume they do collide, we know that the x coordinares are the same, hence 2t = 60 - 4t 6t = 60 t = 10 This means we know that after 10 minutes, the tractors are the same distance from the y axis of the graph (or the west wall of the field). Similarly, if the tractors collide, the y coordinates must be the same, hence we have the same sum... 60 - 4t = 2t 60 = 6t t = 10 Since the tractors are also the same distance from the x axis (south wall of the field) at the same time as they are at the same distance from the y axis (the west wall of the field), they must collide. Therefore, we know that they collide when t = 10 minutes, and from this we can work out that they collide at the point (20,20). As for if they didn't collide and finding the shortest distance, I believe we can actually do this without calculus... The distance between any points (x1,y1) and (x2,y2) on any graph is given by the formula... squareroot( (x1 - x2)^2 + (y1 - y2)^2 ) x1 = 2t x2 = 60-4t y1 = 60-4t y2 = 2t (where t is the time in minutes from the start). Putting these into the formula, we get... squareroot( (2t - (60 - 4t))^2 + (60 - 4t - 2t)^2 ) = squareroot( (2t - 60 + 4t)^2 + (60 - 6t)^2 ) = squareroot( (6t - 60)^2 + (60 - 6t)^2 ) = squareroot( 36t^2 - 720t + 3600 + 3600 - 720t + 36t^2 ) = squareroot( 72t^2 - 1440t + 7200) Since we want to know when this value is at a minimum, we want to minimise this expression... ** Distance = squareroot( 72t^2 - 1440t + 7200 ) ** squaring each side gives... Distance^2 = 72t^2 - 1440t + 7200 Since minimising the distance squared will also minimise the distance, we want to find the value of t which minimises the quadratic equation on the right. I assume you're aware of the method of completing the square with quadratics (if not, let me know), so... 72t^2 - 1440t + 7200 t^2 - 20t + 100 (t - 10)^2 - 100 + 100 (t - 10)^2 Hence the minimum value of this expression occurs when t = 10 (as we found before)...and using t = 10 makes the distance expression above equal to zero...hence the tractors collide! Hope this helps. skuesey. === Subject: Re: Calculus problem > Hey there, > ok, I think this problem is fairly simple, here's what I understand by > the question (please correct me if I'm wrong)... > The square has an area of 3600 sqft, so each side is 60ft long. > So, in a plan view of the field, Big Bertha enters at the top left, and > Tiny Tonka enters at the bottom right. > Now, modelling the plan view mentioned above above as a graph with > point (0,0) at the bottom left corner of the field, we know that after > t minutes, Big Bertha's x coordinate is given by 2t (since she moves 2 > feet east each minute), and the corresponding y coordinate is given by > 60 - 4t (since she starts at the top of the field (hence the 60) and > moves 4 feet south every minute (hence the -4t)). > Similarly, after t minutes, Tiny Tonka's x coordinate is given by 60 - > 4t and the corresponding y co-ordinate is given by 2t. > To prove the two tractors collide, we need to show that for some value > of t, the coordinates of each tractor are the same. > If we assume they do collide, we know that the x coordinares are the > same, hence > 2t = 60 - 4t > 6t = 60 > t = 10 > This means we know that after 10 minutes, the tractors are the same > distance from the y axis of the graph (or the west wall of the field). > Similarly, if the tractors collide, the y coordinates must be the same, > hence we have the same sum... > 60 - 4t = 2t > 60 = 6t > t = 10 > Since the tractors are also the same distance from the x axis (south > wall of the field) at the same time as they are at the same distance > from the y axis (the west wall of the field), they must collide. > Therefore, we know that they collide when t = 10 minutes, and from this > we can work out that they collide at the point (20,20). > As for if they didn't collide and finding the shortest distance, I > believe we can actually do this without calculus... > The distance between any points (x1,y1) and (x2,y2) on any graph is > given by the formula... > squareroot( (x1 - x2)^2 + (y1 - y2)^2 ) > x1 = 2t > x2 = 60-4t > y1 = 60-4t > y2 = 2t > (where t is the time in minutes from the start). Putting these into the > formula, we get... > squareroot( (2t - (60 - 4t))^2 + (60 - 4t - 2t)^2 ) > = squareroot( (2t - 60 + 4t)^2 + (60 - 6t)^2 ) > = squareroot( (6t - 60)^2 + (60 - 6t)^2 ) > = squareroot( 36t^2 - 720t + 3600 + 3600 - 720t + 36t^2 ) > = squareroot( 72t^2 - 1440t + 7200) > Since we want to know when this value is at a minimum, we want to > minimise this expression... > ** Distance = squareroot( 72t^2 - 1440t + 7200 ) ** > squaring each side gives... > Distance^2 = 72t^2 - 1440t + 7200 > Since minimising the distance squared will also minimise the distance, > we want to find the value of t which minimises the quadratic equation > on the right. I assume you're aware of the method of completing the > square with quadratics (if not, let me know), so... > 72t^2 - 1440t + 7200 > t^2 - 20t + 100 > (t - 10)^2 - 100 + 100 > (t - 10)^2 > Hence the minimum value of this expression occurs when t = 10 (as we > found before)...and using t = 10 makes the distance expression above > equal to zero...hence the tractors collide! > Hope this helps. > skuesey. I think I learned more from this post than the sum of anything I learned during my Calculus class this year. I am obviously going to retake this in college, and I am considering retaking Precalculus as depends on this because I haven't done well on most of the tests, making my grade border failure, and this is one of the last few graded items left for the year. === Subject: Re: Calculus problem > Hey there, > ............. >> Hope this helps. > skuesey. Yes skuesey, you solved it but.. what kind of help is that? Solving a homework for a lazy student? Next time think twice what you answer.. a hint (alone) was this question deserved and NOT the full answer. === Subject: Re: Calculus problem > Hey there, > ok, I think this problem is fairly simple, here's what I understand by > the question (please correct me if I'm wrong)... > The square has an area of 3600 sqft, so each side is 60ft long. > So, in a plan view of the field, Big Bertha enters at the top left, and > Tiny Tonka enters at the bottom right. > Now, modelling the plan view mentioned above above as a graph with > point (0,0) at the bottom left corner of the field, we know that after > t minutes, Big Bertha's x coordinate is given by 2t (since she moves 2 > feet east each minute), and the corresponding y coordinate is given by > 60 - 4t (since she starts at the top of the field (hence the 60) and > moves 4 feet south every minute (hence the -4t)). > Similarly, after t minutes, Tiny Tonka's x coordinate is given by 60 - > 4t and the corresponding y co-ordinate is given by 2t. > To prove the two tractors collide, we need to show that for some value > of t, the coordinates of each tractor are the same. > If we assume they do collide, we know that the x coordinares are the > same, hence > 2t = 60 - 4t > 6t = 60 > t = 10 > This means we know that after 10 minutes, the tractors are the same > distance from the y axis of the graph (or the west wall of the field). > Similarly, if the tractors collide, the y coordinates must be the same, > hence we have the same sum... > 60 - 4t = 2t > 60 = 6t > t = 10 > Since the tractors are also the same distance from the x axis (south > wall of the field) at the same time as they are at the same distance > from the y axis (the west wall of the field), they must collide. > Therefore, we know that they collide when t = 10 minutes, and from this > we can work out that they collide at the point (20,20). > As for if they didn't collide and finding the shortest distance, I > believe we can actually do this without calculus... > The distance between any points (x1,y1) and (x2,y2) on any graph is > given by the formula... > squareroot( (x1 - x2)^2 + (y1 - y2)^2 ) > x1 = 2t > x2 = 60-4t > y1 = 60-4t > y2 = 2t > (where t is the time in minutes from the start). Putting these into the > formula, we get... > squareroot( (2t - (60 - 4t))^2 + (60 - 4t - 2t)^2 ) > = squareroot( (2t - 60 + 4t)^2 + (60 - 6t)^2 ) > = squareroot( (6t - 60)^2 + (60 - 6t)^2 ) > = squareroot( 36t^2 - 720t + 3600 + 3600 - 720t + 36t^2 ) > = squareroot( 72t^2 - 1440t + 7200) > Since we want to know when this value is at a minimum, we want to > minimise this expression... > ** Distance = squareroot( 72t^2 - 1440t + 7200 ) ** > squaring each side gives... > Distance^2 = 72t^2 - 1440t + 7200 > Since minimising the distance squared will also minimise the distance, > we want to find the value of t which minimises the quadratic equation > on the right. I assume you're aware of the method of completing the > square with quadratics (if not, let me know), so... > 72t^2 - 1440t + 7200 > t^2 - 20t + 100 > (t - 10)^2 - 100 + 100 > (t - 10)^2 As this is a calculus class I imagine one is meant to use calculus to find the minimum. > Hence the minimum value of this expression occurs when t = 10 (as we > found before)...and using t = 10 makes the distance expression above > equal to zero...hence the tractors collide! > Hope this helps. > skuesey When something is obviously homework I'm not sure it's such a great idea to explain the solution in complete detail. I think a few hints as to the method would be more appropriate. . === Subject: Re: Calculus problem >I need to do this assignment from my Calc class but I have no I idea > what to do with it. Here is the exact text from my worksheet: > Two tractors are plowing through a cornfield. The field is in a shape > of a square with an area of 3600 square feet. The first tractor, Big > Birtha, enters from the left side of the field at 60 feet. The second > tractor, Tiny Tonka, enters from the lower edge at 60 feet. After 1 > minute, the tractors have moved to a new location. Big Birtha has gone > 4 ft South and 2 feet East from its old position. Tiny Tonka has gone 2 > feet north and 4 feet West from its old position. > Question: Will the two tractors collide if they maintain their speeds > and directions? If so, when? If not, how close do they actually come > to each other? > This is designed to be drawn out on a graph so the Big Birtha is > comming from the y-axis and the Tiny Tonka is comming from the x-axis. > If anyone could please explain a process to solve this and any > advance. > NOTE: I know the speeds and size of the field are unrealistic but its > just to apply calc to something slightly real-life I guess. The size > if the tractors is irrelevant for this, just use an exact point. trivial do your homework! === Subject: Re: What does the area under a curve actually mean? > OK, but how did the people FIGURE OUT that position is the area under > the curve of velocity? What made them think position is the area under > the curve of velocity? I really have no idea. Let's consider a person walking down a straight, level, pathway. If the person walked down the path at a constant speed (speed we define by setting a unit length and a unit time) then we know that if it takes him/her to walk 1meter in 1 sec then in general, by symmetry, it will take him/her x sec to walk x meters. Trivial no? Suppose that I have attached a radar gun behind the person so that I can measure his/her speed constantly. You can visualize that I will attain a graph/plot of the person's velocity as a function of time. Do I have any way of telling you how far the person went? Or even where he/she was at any given time? (the Uncertainty Principle would say no, but let's just say YES for now). How? Recall that when the person's velocity was constant, we could have simply multiplied his/her speed by the time it took to cover a certain distance. We can (falsely) assume that if we cut the trip up into intervals of equal lengths of time (make a table, whatever) and for each interval assume a constant velocity, then we can use the simple case and sum all of the distances up to get a 'pretty good idea' of 'where and when', etc. (You may encounter the name for this process; it is called a Lower Riemann Integral.) That's it. We went from the simple to estimating for a case that was more complex. In introductory Calculus texts you may find the proofs of how Differentiation and Integration are connected (by the FTC(s)). If you go on, you may find it more convenient to say that integration is the more natural counterpart to these two (and so you may say differentiation is the inverse of integration!). Think about it next time you are walking down the street. You don't need any calculus training to figure out that it makes sense. Good luck. M>Now, I honestly don't know where to start in proving that there is no >such set P in complex numbers to do that. Another poster already gave you a hint. Is i in P? What is i*i? is -i in P? What is (-i)*(-i)? However you answer those questions, you get a contradiction, so there is no P satisfying the conditions. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: > Suppose F is a collection of numbers, and there is a subset of F, > called P which is closed under addition and miltiplication, so that for > each f in F exactly one of three this is true: 1) f ü P 2) -f ü P > 3) f=0. Then we say that F is ordered. That's not the definition I learned. It looks like a definition of postive numbers to me. Check out http://mathworld.wolfram.com/PartialOrder.html http://mathworld.wolfram.com/TotallyOrderedSet.html for more suitable definitions. --- Christopher Heckman === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: > > Suppose F is a collection of numbers, and there is a subset of F, > called P which is closed under addition and miltiplication, so that for > each f in F exactly one of three this is true: 1) f ü P 2) -f ü P > 3) f=0. Then we say that F is ordered. > That's not the definition I learned. It looks like a definition of > postive numbers to me. > Check out > http://mathworld.wolfram.com/PartialOrder.html > http://mathworld.wolfram.com/TotallyOrderedSet.html > for more suitable definitions. > She is defining ordered field, not ordered set. Even so, this still doesn't look like the definition I used ... The definition of an ordered field I learned was: (F,+,*,<=) is an ordered field if (1) (F, +, *) is a field) (2) <= is a total order (3) If a <= b, then a+c <= b+c, for all a,b,c (4) If a <= b and 0 <= c, then a*c <= b*c, for all a,b,c. The definition she is using may be equivalent (when a field is ordered), but it still seems strange to call it the definition. --- Christopher Heckman === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: > > Suppose F is a collection of numbers, and there is a subset of F, > called P which is closed under addition and miltiplication, so that for > each f in F exactly one of three this is true: 1) f ü P 2) -f ü P > 3) f=0. Then we say that F is ordered. > > That's not the definition I learned. It looks like a definition of > postive numbers to me. > > Check out > > http://mathworld.wolfram.com/PartialOrder.html > http://mathworld.wolfram.com/TotallyOrderedSet.html > > for more suitable definitions. > She is defining ordered field, not ordered set. > Even so, this still doesn't look like the definition I used ... The > definition of an ordered field I learned was: (F,+,*,<=) is an ordered > field if > (1) (F, +, *) is a field) > (2) <= is a total order > (3) If a <= b, then a+c <= b+c, for all a,b,c > (4) If a <= b and 0 <= c, then a*c <= b*c, for all a,b,c. > The definition she is using may be equivalent (when a field is > ordered), but it still seems strange to call it the definition. > --- Christopher Heckman They are equivalent, and I'm guessing she is calling it the definition because that's what was provided in a classroom context. Under your definition, the set: P = { a | 0 <= a & a nonzero } satisfies the requirements of her definition. Given such a positive cone, closed under + and *, it is possible to define a <= b as either a = b or b - a belongs to P, and then to show each of (2),(3),(4). For example, in (3) we have two cases for a <= b: a = b implies a + c = b + c Otherwise b - a = (b+c) - (a+c) is in P, so a + c <= b + c. (4) is similar, using the same alternatives. When b - a is in P and c is in P, then (b-a)*c is in P by the properties of her definition, and a*c <= b*c. A positive cone, e.g. closed under addition, does not generally induce a total order. But here the germ of trichotomy is in the assumption that exactly one holds, for every a: a is in P OR a = 0 OR -a is in P -- chip === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: > > Suppose F is a collection of numbers, and there is a subset of F, > called P which is closed under addition and miltiplication, so that for > each f in F exactly one of three this is true: 1) f ü P 2) -f ü P > 3) f=0. Then we say that F is ordered. > > That's not the definition I learned. It looks like a definition of > postive numbers to me. > > Check out > > http://mathworld.wolfram.com/PartialOrder.html > http://mathworld.wolfram.com/TotallyOrderedSet.html > > for more suitable definitions. > > She is defining ordered field, not ordered set. > Even so, this still doesn't look like the definition I used ... The > definition of an ordered field I learned was: (F,+,*,<=) is an ordered > field if > (1) (F, +, *) is a field) > (2) <= is a total order > (3) If a <= b, then a+c <= b+c, for all a,b,c > (4) If a <= b and 0 <= c, then a*c <= b*c, for all a,b,c. > The definition she is using may be equivalent (when a field is > ordered), but it still seems strange to call it the definition. > They are equivalent, Yes, that was something I worked through while going to sleep last night. > and I'm guessing she is calling it > the definition because that's what was provided in a > classroom context. Under your definition, the set: > P = { a | 0 <= a & a nonzero } > satisfies the requirements of her definition. Given > such a positive cone, closed under + and *, it is > possible to define a <= b as either a = b or b - a > belongs to P, and then to show each of (2),(3),(4). > For example, in (3) we have two cases for a <= b: > a = b implies a + c = b + c > Otherwise b - a = (b+c) - (a+c) is in P, so > a + c <= b + c. > (4) is similar, using the same alternatives. When > b - a is in P and c is in P, then (b-a)*c is in P by > the properties of her definition, and a*c <= b*c. > A positive cone, e.g. closed under addition, does > not generally induce a total order. But here the > germ of trichotomy is in the assumption that > exactly one holds, for every a: > a is in P OR a = 0 OR -a is in P That was the part I was wondering about. --- Christopher Heckman === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: The complex field can be ordered in many ways, but none of those orderings are compatible with the 'ordered ring' requirements below. Suppose (F,+,*) is a ring and there is a subset of F, called P which is closed under the addition and multiplication of the ring , so that for each f in F exactly one of three this is true: 1) f is in P 2) the additive inverse of f is in P 3) f is the additive identity of the ring. Then we say that F is an ordered ring. > Now, I honestly don't know where to start in proving that there is no > such set P in complex numbers to do that. > Any help you can provide would be wonderful. Show that if x^2 < 0 neither x nor -x can be in P === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: > Suppose F is a collection of numbers, and there is a subset of F, > called P which is closed under addition and miltiplication, so that for > each f in F exactly one of three this is true: 1) f ü P 2) -f ü P > 3) f=0. Then we say that F is ordered. > Now, I honestly don't know where to start in proving that there is no > such set P in complex numbers to do that. If there were such a P, would i or -i be in it? Remember P is closed under addition and multiplication. === Subject: Re: Can anyone help me prove that the complex field is not ordered? > If there were such a P, would i or -i be in it? > Remember P is closed under addition and multiplication. That's what I'm having trouble with. I understand what all the posts are saying, but it's that next step that I'm having trouble with. Maybe I'm misunderstanding this but would the following be valid? Suppose F = all complex numbers, and P = a + bi where a,b are positive. For f = m + ni, where m, n are positive, condition 2) is violated because -f = -m - ni which violates the set P. Or am I completely off-track? I had never really thought about the orderd-ness of complex numbers...I had just always known that they're not ordered but never knew why. === Subject: Re: Can anyone help me prove that the complex field is not ordered? > If there were such a P, would i or -i be in it? > Remember P is closed under addition and multiplication. > That's what I'm having trouble with. I understand what all the posts > are saying, but it's that next step that I'm having trouble with. You didn't show any work on that next step, but abandoned the essentially identical hints three of us gave. Are you sure you understand what all the posts are saying? > Maybe I'm misunderstanding this but would the following be valid? > Suppose F = all complex numbers, and P = a + bi where a,b are positive. > For f = m + ni, where m, n are positive, condition 2) is violated > because -f = -m - ni which violates the set P. > Or am I completely off-track? I had never really thought about the > orderd-ness of complex numbers...I had just always known that they're > not ordered but never knew why. Yes, you are completely off-track. Please don't take that as an insult. In order to prove the complex field can't be an ordered field, it's not enough to choose one specific set P and show that it doesn't work. You have to show that *no* P can possibly work. Welcome to the precision and abstractness of mathematics. It can take a while to get used to the language and the way of thinking, but it's very rewarding. You need to make sure you understand all the pieces of the problem that you have been given. What does it mean to say that a set P is closed under addition and multiplication? What do the three conditions you listed mean? I'm certain So before you go any further you need to make sure you understand just what exactly an ordered field is. It's not enough to have a sort of intuitive feel for it; you need to grasp the definition. Then you can work on showing that the complex field can't be made to satisfy this definition. === Subject: Re: Can anyone help me prove that the complex field is not ordered? Ok, then I guess I don't understand the pieces. It would be great if you could help me in understanding it because I really do want to understand where the non-order-ness of complex numbers comes from. I've never worked this in-depth with complex numbers. means that if you have two numbers, x and y that are in P, that x+y is in P and that x*y is in P. This is the question that was posed, more or less: In general,a field F is called an ordered field, provided that it contains a positive subset P with respect to which the following axioms hold. (a) If x is in F, then one and only one of the following is true: x=0, x is in P, or -x is in P (b) If x and y are elements in P, then so is their product xy. (c) If x and y are elements of P, then so is their sum x + y. Prove that C is not an ordered field. I'm trying to understand each of the axioms. Would it be enough to say: If ai > 0 and bi > 0, then to satisfy (b) condition, (ai)(bi) should be > 0, but are not since (ai)(bi) = -ab which is < 0. Please, I really do want to understand this concept and I greatly appreciate the help. === Subject: Re: Can anyone help me prove that the complex field is not ordered? Ok, I did some research on the axioms and I think I finally understand it. Complex numbers cannot be ordered because if complex numbers were ordered, the following would be true: i < 0 or i = 0 or i > 0 None of the relationships is acceptable because: a) Suppose i < 0, then -i > 0, and then the closed under multiplication portion is violated because (-i)(-i) > 0 is false because it gives that -1 > 0. b) Suppose i = 0. Then the laws of order give (i)(i) = 0. But this is false because i^2 = -1. c) Suppose i > 0. Then we have (i)(i) > 0, which again gives -1 > 0. This is false. === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Ok, I did some research on the axioms and I think I finally understand > it. > Complex numbers cannot be ordered because if complex numbers were > ordered, the following would be true: > i < 0 or i = 0 or i > 0 > None of the relationships is acceptable because: > a) Suppose i < 0, then -i > 0, and then the closed under > multiplication portion is violated because (-i)(-i) > 0 is false > because it gives that -1 > 0. > b) Suppose i = 0. Then the laws of order give (i)(i) = 0. But this is > false because i^2 = -1. > c) Suppose i > 0. Then we have (i)(i) > 0, which again gives -1 > 0. > This is false. You need to go back to basics a bit more. Rather than talking about i < 0 or i = 0 or i > 0, phrase your argument in terms of P, using the definition of ordered field. Exactly one holds: i is in P or i = 0 or -i is in P Get a contradiction from this. (Hint: i is not zero. Work on the other two cases.) -- c === Subject: Re: Can anyone help me prove that the complex field is not ordered? > In order to prove the complex field can't be an ordered field, it's > not enough to choose one specific set P and show that it doesn't > work. You have to show that *no* P can possibly work. > Welcome to the precision and abstractness of mathematics. > It can take a while to get used to the language and the way of > thinking, but it's very rewarding. > You need to make sure you understand all the pieces of the > problem that you have been given. What does it mean to say > that a set P is closed under addition and multiplication? > What do the three conditions you listed mean? I'm certain > So before you go any further you need to make sure you > understand just what exactly an ordered field is. It's not > enough to have a sort of intuitive feel for it; you need to > grasp the definition. Then you can work on showing that > the complex field can't be made to satisfy this definition. Ok, then I guess I don't understand the pieces. It would be great if you could help me in understanding it because I really do want to understand where the non-order-ness of complex numbers comes from. I've never worked this in-depth with complex numbers. means that if you have two numbers, x and y that are in P, that x+y is in P and that x*y is in P. Please, I really do want to understand this concept and I greatly appreciate the help. === Subject: Re: Can anyone help me prove that the complex field is not ordered? > Hi everyone, > I'm having trouble starting the proof that the complex field is not > ordered. I understand the definition of order: > Suppose F is a collection of numbers, and there is a subset of F, > called P which is closed under addition and miltiplication, so that for > each f in F exactly one of three this is true: 1) f ü P 2) -f ü P > 3) f=0. Then we say that F is ordered. > Now, I honestly don't know where to start in proving that there is no > such set P in complex numbers to do that. > Any help you can provide would be wonderful. Pick a favorite nonzero imaginary number and ask yourself: Is it in P, or is its additive inverse in P? Find a contradiction. -- c === Subject: Re: Can anyone help me prove that the complex field is not ordered? >Hi everyone, >I'm having trouble starting the proof that the complex field is not >ordered. I understand the definition of order: >Suppose F is a collection of numbers, and there is a subset of F, >called P which is closed under addition and miltiplication, so that for >each f in F exactly one of three this is true: 1) f ? P 2) -f ? P >3) f=0. Then we say that F is ordered. >Now, I honestly don't know where to start in proving that there is no >such set P in complex numbers to do that. Consider the number i. You will need some additional results you may or may not already have proven. === Subject: Re: ? is this e-problem >> eigenvalue problem: A*x = lamda*x => find lamda and x >> generalized e-probelm: A*x = lamba*B*x => find lamba and x >> How about A*x = b*B*x+c*C*x => find b, c, and x? >> Even more: A*x = sum( b(n)*B_n, n = 1 to N )*x? >... > assuming that the matrices are n*n, since you have one more variable > than equations (you should implicitely add some scaling equation > |x|=1), one would expect curves of eigenvalues: For example if > lambda_1 (lambda_2) is an eigenvalue of the problem A*x = lambda B*x > (A*x = lambda C*x), then bc=(lambda_1, 0) and (0, lambda_2) would be > solutions of your equation. There should be curves of solutions > passing through these special ones. > But beware: These curves need not be connected: Generalized > eigenproblems behave differently, if B becomes singular. Looks like a clear understanding of A*x = lambda*B*x is the key to understand cases involving more terms. Let's do this first. (1) A*x = lamba*B*x while B is singular, => x and lambda = ? (2) this time A is singular, => x and lambda = ? (3) Both A and B are regular, but A and B have two eigenvalues the same, => x and lambda = ? by Cheng Cosine Jun/09/2k6 NC === Subject: Re: Cantorian pseudomathematics <16048404.1149434840703.JavaMail.jakarta@nitrogen.mathforum.org> <448567aa$0$2026$ba620dc5@text.nova.planet.nl> No doubt Cantor's work is extremely interesting and beautiful >> and I am quite nervous of the prospect of losing his `paradise', >> but in face of so many paradoxes (see some of the ones Jaynes >The original manuscript is available online. >http://omega.albany.edu:8008/JaynesBook.html > All of these paradoxes are well-known and unavoidable. > Looked at properly, some of them are in some way paradoxical, > but others are not. FWIW, presumably, when looked at properly, no paradox is truly paradoxical. I don't think Jaynes was claiming that he had found inconsistencies in set theory, but rather, that a naive but typical attempt to apply set theory to probability theory leads to incorrect results and apparent inconsistencies. === Subject: Re: Cantorian pseudomathematics <16048404.1149434840703.JavaMail.jakarta@nitrogen.mathforum.org> <448567aa$0$2026$ba620dc5@text.nova.planet.nl> naive person would want? apparent. The author discusses how infinite sets should be used in the context of probability theory, which seems to have no bearing on the question in general. === Subject: A place for the curious, critical and bored. To whom it may concern, teh water cooler.com - Is a place for the curious, critical and bored - Discuss, debate anything ranging from politics, religion, science, current events, music, philosophy - teh water cooler.com We are looking for intelligent members who would be willing too moderate discussions or join in current ones. www.tehwatercooler.com (intentional misspelling) http://www.tehwatercooler.com === Subject: Re: A place for the curious, critical and bored. > .... > We are looking for intelligent members who would be willing too > moderate discussions or join in current ones.... ^^^ Should they be clever enough to know the difference between to and too? === Subject: Re: A place for the curious, critical and bored. We are looking for intelligent members who would be willing too > moderate discussions or join in current ones.... ^^^ > Should they be clever enough to know the difference between to > and too? AND bored enough to point it out. cordially Y.T. -- Remove YourClothes before you email me. === Subject: Re: A place for the curious, critical and bored. > To whom it may concern, > teh water cooler.com - Is a place for the curious, critical and bored - > Discuss, debate anything ranging from politics, religion, science, > current events, music, philosophy - teh water cooler.com > We are looking for intelligent members who would be willing too > moderate discussions or join in current ones. Intelligent people are never bored, because they always have work to do. -- Ioannis === Subject: Re: route planners Hey there, A few suggestions for some algorithms and terms that you may want to i) Dijkstra's algorithm - finds the shortest distance between two nodes of a graph (as mentioned in the previous post) ii) Prim's algorithm/Kruskal's algorithm - finds the minimum amount of road needed to connect nodes iii) Route inspection algorithm - finds the total distance you must travel if you must traverse each road at least once. iv) Nearest neighbour algorithm/Travelling salesperson problem - finds a solution (but not necessarily the best solution) to the problem of visiting every node at least once and returning to the starting node If you're wondering how I know all these, I am currently studying discrete maths at high school and have to learn these algorithms for my exam in a few days time. Hope this helps, skuesey. === Subject: Re: route planners > Hey there, > A few suggestions for some algorithms and terms that you may want to > i) Dijkstra's algorithm - finds the shortest distance between two nodes > of a graph (as mentioned in the previous post) > ii) Prim's algorithm/Kruskal's algorithm - finds the minimum amount of > road needed to connect nodes > iii) Route inspection algorithm - finds the total distance you must > travel if you must traverse each road at least once. > iv) Nearest neighbour algorithm/Travelling salesperson problem - finds > a solution (but not necessarily the best solution) to the problem of > visiting every node at least once and returning to the starting node > If you're wondering how I know all these, I am currently studying > discrete maths at high school and have to learn these algorithms for my > exam in a few days time. > Hope this helps, > skuesey. Thanx a LOT ! :-) martijn === Subject: Re: route planners > I am trying to grasp the essence of the working/algorithm of routeplanners > (like road routeplanners on internet or for Tom Tom's, or railroad > routeplanners, etc) There are many different types of route planning problems. Some of them are very easy, and some are almost impossibly difficult. For example, if you have a given network, with arcs connecting the nodes (like streets connecting locations in a city), and if the time or distance along each arc is known, then it is very easy to work out a shortest or fastest route from point A to point B. For example, look up Dykstra's Algorithm (although that may be misspelled). Airlines often work with complex models of routes in order to minimize fuel costs, for example, given passenger demands on various legs and varying fuel prices at different locations. UPS and FEDEX need to plan routes and locations of depots, etc. Trucking companies need to deliver goods to customers, and would like to either minimize fuel cost, or use the smallest number of trucks, etc. Problems of this latter type are very difficult to solve exactly, so heuristsics may be used in real-world, industrial-strength problems. Do a Google search on the vehicle routing problem. Finally, you may have more luck getting replies from the operations research newsgroup 'sci.op-research', since many of the folks who develop the algorithms for such problems consider themselves to be operations researchers rather than (applied) mathematicians. Also, you may find more info in the INFORMS website www.informs.org. Good luck R.G. Vickson > Perhaps someone could point me to some basic theory about this subject which > could clear this 'mystery'. > I myself am not really a hardcoremathematician, although I have studied > physics. So hopefully the backgrounds are not too mathematic-theoretic. > martijn === Subject: Re: route planners > I am trying to grasp the essence of the working/algorithm of routeplanners > (like road routeplanners on internet or for Tom Tom's, or railroad > routeplanners, etc) > There are many different types of route planning problems. Some of > them are very easy, and some are almost impossibly difficult. For > example, if you have a given network, with arcs connecting the nodes > (like streets connecting locations in a city), and if the time or > distance along each arc is known, then it is very easy to work out a > shortest or fastest route from point A to point B. For example, look up > Dykstra's Algorithm (although that may be misspelled). Sorry: it IS misspelled. Do a Google search of Dijkstra's Algorithm. If you do it on Dykstra's Algorithm, you will be led off in another, totally different direction! RGV > Airlines often > work with complex models of routes in order to minimize fuel costs, for > example, given passenger demands on various legs and varying fuel > prices at different locations. UPS and FEDEX need to plan routes and > locations of depots, etc. Trucking companies need to deliver goods to > customers, and would like to either minimize fuel cost, or use the > smallest number of trucks, etc. Problems of this latter type are very > difficult to solve exactly, so heuristsics may be used in real-world, > industrial-strength problems. Do a Google search on the vehicle > routing problem. Finally, you may have more luck getting replies from > the operations research newsgroup 'sci.op-research', since many of the > folks who develop the algorithms for such problems consider themselves > to be operations researchers rather than (applied) mathematicians. > Also, you may find more info in the INFORMS website www.informs.org. > Good luck > R.G. Vickson > Perhaps someone could point me to some basic theory about this subject which > could clear this 'mystery'. > I myself am not really a hardcoremathematician, although I have studied > physics. So hopefully the backgrounds are not too mathematic-theoretic. > martijn === Subject: Re: An integer sequence <6397029.1149857480503.JavaMail.jakarta@nitrogen.mathforum.org > Let (a_n)_{n from N } be a sequence such that > pa_{n+2} = qa_{n+1} + ra_n for integer p,q,r such > that gcd(q,r) isn't divisible by p. It is known that > all the a_n'as are integers. Prove that (a_n)_n is a > geometric progression. > > What are you ideas for this problem? > Solve the characteristic difference equation. > Express a_n as a linear > combination > of powers of its roots. > px^2 = qx + r. > X_n = C_1* X_1^n + C_2* X_2^n where > X_1 and X_2 are the roots of the equation px^2 - qx -r = 0 > What i dont' know is how can i proof that if > C_1* X_1^n + C_2* X_2^n is integer for all natural n, then > r = 0. > That's the point where i'm stuck If you can show C_1 = 0 or C_2 = 0, you'd be done. --- Christopher Heckman === Subject: Re: An integer sequence <6397029.1149857480503.JavaMail.jakarta@nitrogen.mathforum.org > Let (a_n)_{n from N } be a sequence such that > pa_{n+2} = qa_{n+1} + ra_n for integer p,q,r such > that gcd(q,r) isn't divisible by p. It is known that > all the a_n'as are integers. Prove that (a_n)_n is a > geometric progression. > > What are you ideas for this problem? > > Solve the characteristic difference equation. > Express a_n as a linear > combination > of powers of its roots. > > px^2 = qx + r. > X_n = C_1* X_1^n + C_2* X_2^n where > X_1 and X_2 are the roots of the equation px^2 - qx -r = 0 > What i dont' know is how can i proof that if > C_1* X_1^n + C_2* X_2^n is integer for all natural n, then > r = 0. > That's the point where i'm stuck > If you can show C_1 = 0 or C_2 = 0, you'd be done. Doesn't this imply that at least one of the roots is an integer ? pa(n+2) = q*a(n+1) +r*a(n) pa(n+1) = q*a(n) +r*a(n-1) q*a(n+1) +r*a(n) = 0 mod p q*a(n) +r*a(n-1) = 0 mod p => a(n+1)*a(n-1) - a(n)^2 = 0 mod p => a0, a0*k, a0*k^2, ...,a0* k^n , .. )mod p pk^2 -qk - r = 0 mod p but pk^2 -qk - r = 0 => k integer root === Subject: Re: An integer sequence > Let (a_n)_{n from N } be a sequence such that > pa_{n+2} = qa_{n+1} + ra_n for integer p,q,r such that gcd(q,r) isn't divisible by p. > It is known that all the a_n'as are integers. Prove that (a_n)_n is a geometric progression. > What are you ideas for this problem? Hmmm... A surprising result, which I first thought was not true. Define R_n = a_{n+1} / a_n; you need to show that R_n = R_{n+1} for all n. Dividing your equation through by a_{n+1}, you get p a_{n+2} / a_{n+1} = q + r a_n / a_{n+1}, or p R_{n+1} = q + r / R_n. Thus R_{n+1} = (q + r / R_n) / p, which looks strange, but maybe you can manipulate this some more ... --- Christopher Heckman === Subject: Re: bayesians versus frequentists >>BTW - I *do* think that Bayesian ideas have a good place! >>Namely, in subjective circumstances. >> I want to know when frequentist ideas ever have a good place. >> It seems to me that it is only for simplicity; the frequentist >> keeps track of fewer numbers. >The frequentist view fits into the currently fashionable set-theoretic >foundational framework for mathematics better than the Bayesian view, >and hence, the pure mathematicians favor it. >I have argued in the past that a better foundational framework for >mathematics would use some notion of fuzzy sets, and the Bayesian >view would then be seen as having a superior fit to mathematics. to probability to form a viable adjunct. It seems we need a Boolean algebra for a truth value system. Probability is NOT a truth value in the logical sense, as the truth value of a Boolean combination can only depend on the truth values of the components. Let P(A) = .5; then P(~A) = .5. So what is P(A&A) and P(A&~A)? Understanding basic set theory and logic helps in avoiding these problems. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists >> One of the conclusions from decision theory is that >> so-called objective procedures can often be very bad, in >> the sense that they can be even uniformly improved in many >> cases. All good procedures (in that sense) are limits of >> Bayes, but limits of Bayes are not always good. >I'd like to understand that last comment better. Could you expound a >little on that or give us a reference? The classic example is that of X being normal(m, 1). Let m have density proportional to exp(m - q*m^2), q small. The limit of the Bayes procedure is to give m the limit density, and the estimate with this prior is X+1, which is uniformly worse than X. For estimation, any peculiar prior is a limit of proper priors, and if the appropriate integral approximations are valid, the Bayes procedure is likewise the limit. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists iD8DBQFEige4vmGe70vHPUMRAhqvAJ9j5J7jmPaGqiVVJZnxRUNlhR8uaACfZPsl QyfWr1DIvHvgkW8O8mtT34Y= =J8kj >dave says... >>The frequentist view fits into the currently fashionable set-theoretic >>foundational framework for mathematics better than the Bayesian view, >>and hence, the pure mathematicians favor it. >I don't see that at all. I would think that a pure mathematician would >reject frequentism as nonrigorous. The probability of X, according to >frequentism, is the limit (number of instances of X)/(number of trials) >as the number of trials goes to infinity. But mathematically speaking, >there is no reason for this ratio to have a limit. Yes, exactly. I always thought the important dispute was between Bayesian methods vs. Neyman-Pearson methods. This has to do with experimental methodology, mostly in the social sciences. >The pure mathematician is more likely to define probabilities axiomatically: >A probability distribution on a set E is a partial function from the power >set of E to [0,1] such that blah, blah, blah. (Where blah blah includes >countable additivity, etc.) If the concern is with a philosophical basis, then I agree that the axiomatic definition is appropriate. Neither Bayesian nor frequentist philosophies are mathematically satisfying. === Subject: Re: bayesians versus frequentists >>dave says... >The frequentist view fits into the currently fashionable set-theoretic >foundational framework for mathematics better than the Bayesian view, >and hence, the pure mathematicians favor it. >>I don't see that at all. I would think that a pure mathematician would >>reject frequentism as nonrigorous. The probability of X, according to >>frequentism, is the limit (number of instances of X)/(number of trials) >>as the number of trials goes to infinity. But mathematically speaking, >>there is no reason for this ratio to have a limit. >Yes, exactly. >I always thought the important dispute was between Bayesian methods vs. >Neyman-Pearson methods. This has to do with experimental methodology, >mostly in the social sciences. No; Bayesian methods are the extension of Neyman-Pearson methods using consistency. The Neyman-Peason Lemma states that all reasonable 2-decision problems with 2 states of nature are Bayesian. Add the nature isn't against you principle and it extends; add the risk principle, and now you have full Bayesian, at least for finite numbers of states of nature. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists I haven't been following along, but I thought more than two choices existed for conceptualizing probability. As far as pure mathematics goes, can all the concepts offered be axiomized by Kolmogorov's (sp?) scheme? Relative frequentists think of probabilities as frequencies that exist independently of any observer. Bayesians, I guess, think of probabilities as conditions for relative consistency in the minds of observer. (No dutch book.) Different observers may have different probabilities, and each observer could be internally consistent. But isn't there a conception of probability as being like logic (Keynes)? Any observer, if he is thinking logically, according to this concept should come to the same conclusion. And what is fiduciary probability, if I recall correctly? -- Whether strength of body or of mind, or wisdom, or virtue, are found in proportion to the power or wealth of a man is a question fit perhaps to be discussed by slaves in the hearing of their masters, but highly unbecoming to reasonable and free men in search of the truth. -- Jean Jacques Rousseau === Subject: Re: bayesians versus frequentists >I haven't been following along, but I thought more than two choices existed >for conceptualizing probability. >As far as pure mathematics goes, can all the concepts offered be axiomized >by Kolmogorov's (sp?) scheme? Not quite, but close. Instead of DEFINING probability in terms of subsets of a measure space, and random variables as functions, if one postulates that in any situation one can REPRESENT the problem in this manner, with possibly many choices of spaces available, it becomes adequate. >Relative frequentists think of probabilities as frequencies that exist >independently of any observer. They seem to, but relative frequencies require independence (only defined in probability) and identical conditions, which never exist. >Bayesians, I guess, think of probabilities as conditions for relative >consistency in the minds of observer. (No dutch book.) Different observers >may have different probabilities, and each observer could be internally >consistent. Some do, but not all. One can consider the utility not knowing the state of nature as an integral over those states of the utilities given those states. The belief Bayesian usually considers the utilities to be fixed, and the weights as priors, but there seems to be no way to compare utilities for different states of nature, and it is certainly the case that the action taken only depends on the product. This is the conclusion from coherence, which is acting in a self-consistent manner. >But isn't there a conception of probability as being like logic (Keynes)? >Any observer, if he is thinking logically, according to this concept should >come to the same conclusion. The representation above uses the Boolean algebra of measurable sets as a truth-value space. >And what is fiduciary probability, if I recall correctly? Something invented by Fisher to attempt to avoid having to be Bayesian. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists <610635.KAPRJULH@news.dreamscape.com> What about the use of bayesian probabilities in the theory of war and peace? Can we assume that Kim Jung IL and other dictators and terrorists behave rationally according to their Bayesian priors about our response to given actions? >I haven't been following along, but I thought more than two choices existed >for conceptualizing probability. >As far as pure mathematics goes, can all the concepts offered be axiomized >by Kolmogorov's (sp?) scheme? > Not quite, but close. Instead of DEFINING probability in > terms of subsets of a measure space, and random variables > as functions, if one postulates that in any situation one > can REPRESENT the problem in this manner, with possibly > many choices of spaces available, it becomes adequate. >Relative frequentists think of probabilities as frequencies that exist >independently of any observer. > They seem to, but relative frequencies require independence > (only defined in probability) and identical conditions, which > never exist. >Bayesians, I guess, think of probabilities as conditions for relative >consistency in the minds of observer. (No dutch book.) Different observers >may have different probabilities, and each observer could be internally >consistent. > Some do, but not all. One can consider the utility not > knowing the state of nature as an integral over those > states of the utilities given those states. The belief > Bayesian usually considers the utilities to be fixed, and > the weights as priors, but there seems to be no way to > compare utilities for different states of nature, and it > is certainly the case that the action taken only depends > on the product. This is the conclusion from coherence, > which is acting in a self-consistent manner. >But isn't there a conception of probability as being like logic (Keynes)? >Any observer, if he is thinking logically, according to this concept should >come to the same conclusion. > The representation above uses the Boolean algebra of > measurable sets as a truth-value space. >And what is fiduciary probability, if I recall correctly? > Something invented by Fisher to attempt to avoid having > to be Bayesian. > -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists <610635.KAPRJULH@news.dreamscape.com> peace? Can we assume that Kim Jung IL and other dictators and > terrorists behave rationally according to their Bayesian priors about > our response to given actions? Sometimes the rational thing to do is to gamble (i.e. allow you actions to be determined by the throw of the dice). If your actions are predictable, your opponent can use that predictability against you. The point here is that behaving rationally is not the same as behaving predictably. === Subject: Re: bayesians versus frequentists >What about the use of bayesian probabilities in the theory of war and >peace? Can we assume that Kim Jung IL and other dictators and >terrorists behave rationally according to their Bayesian priors about >our response to given actions? Nobody can manage to behave rationally according to his values and priors; the computational problems are unsolvable. Also, in a non-zero sum game, the mechanics of rational behavior are not worked out. Further, can we expect dictators and terrorists to be that rational? The answer seems clearly no. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists > I haven't been following along, but I thought more than two choices existed > for conceptualizing probability. > As far as pure mathematics goes, can all the concepts offered be axiomized > by Kolmogorov's (sp?) scheme? > Relative frequentists think of probabilities as frequencies that exist > independently of any observer. > Bayesians, I guess, think of probabilities as conditions for relative > consistency in the minds of observer. (No dutch book.) Different observers > may have different probabilities, and each observer could be internally > consistent. > But isn't there a conception of probability as being like logic (Keynes)? > Any observer, if he is thinking logically, according to this concept should > come to the same conclusion. > And what is fiduciary probability, if I recall correctly? Which bit of that did Herman Rubin write? Please learn to quote properly. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: bayesians versus frequentists Only heat matters. Light, like dung, is uninteresting by product of heat. > Just to put some people out of their misery... in the unlikely event > that anyone is still interested in what I might have to say. > There is a huge amount more that could be said. > However, attitudes have already hardened, and there seems little point > to continuing with a wordy debate, which is obviously going to generate > a lot more heat than light. > So I will not be writing any more to this thread. > B T === Subject: Re: bayesians versus frequentists <090620061337082328%chenrich@monmouth.com> In other words, an academic mathematician holds a subjective probabilty that his theorems hold interest independently of whether than can be applied to the real world or not. This means that to be a mathematician, you have to be a bayesian who believes in your own subjective evaluation of what is interesting mathematics. > dave says... >The frequentist view fits into the currently fashionable set-theoretic >foundational framework for mathematics better than the Bayesian view, >and hence, the pure mathematicians favor it. > I don't see that at all. I would think that a pure mathematician would > reject frequentism as nonrigorous. The probability of X, according to > frequentism, is the limit (number of instances of X)/(number of trials) > as the number of trials goes to infinity. But mathematically speaking, > there is no reason for this ratio to have a limit. > The pure mathematician is more likely to define probabilities axiomatically: > A probability distribution on a set E is a partial function from the power > set of E to [0,1] such that blah, blah, blah. (Where blah blah includes > countable additivity, etc.) > To the pure mathematician, this is a satisfactory definition of a topic > in pure mathematics, within which there are plenty of problems to > solve, theorems to prove, and so on. The debate starts when people try > to apply the mathematics to non-mathematical problems. > For some time I was of the opinion that applications of probability > theory tended to be of two kinds, and the difference between > frequentists and Bayesians was that they were interested in > different applications. The different kinds of applications are: > 1. Statistical mechanics. Here, it is quite impossible to track > trajectories of a dynamical system with 10^24 degrees of freedom, but > if we replace the individual trajectories by a suitably chosen measure > on configuration space, we get something that is mathematically > tractable and useful. > 2. Insurance. Here, we cannot predict which building will have a fire, > but in a large enough ensmble of buildings we can predict, closely > enough, how many will have fires. > 3. Estimation theory. Here, probabilites are quantitative expressions > of imperfect knowledge. > (Wait. There are three different kinds of applications: quantum > mechanics; statistical mechanics... I mean, /four/ kinds: quantum > mechanics, statistical mechanics, insurance, estimation, and fanatical > devotion to the Pope.) > In my list, insurance is pretty clearly a frequentist application, and > estimation is a Bayesian application. I am not sure whether > statistical machanics is frequentist or Bayesian. > To have a debate about frequentism versus Bayesianism, the debaters > must agree that the question, What is the meaning of proability > theory? has one answer which is both all-inclusive and all-exclusive. > It has to be inclusive of all possible uses of probability theory, and > it has to be exclusive of all alternative meanings. > -- > Chris Henrich > http://www.mathinteract.com > God just doesn't fit inside a single religion. === Subject: Re: bayesians versus frequentists >> If is meaningful to ask what is the probability that Bush will be >> re-elected in the 2008 election. A meaningful and operational >> value of p can be assigned to that future event, and it's likely >> that everyone will have a slightly different value. That is the >> crux of SUBJECTIVE probability. >That is entirely correct. But it all depends on the conditional: >IF it is meaningful to ask.... I don't think it IS meaningful, >except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot >be spoken of (IMHO) in a discussion between two or more people. >The utter failure of Bayesians to extend their methods to multi-person >or (better still) public conclusions, in spite of decades >of failed attempts, is highly significant. There is an excellent reason; it cannot be done. There is no method of arriving at a group decision if the individuals and the group are both self-consistent other than having one member decide. Arrow proved this in his thesis, but did not include randomization axioms. In my 1987 paper in _Statistics and Decisions_, I pointed out that the problem to get self-consistent group actions out of self-consistent actions by the individual had to be Bayesian, in the sense that the utility functions of the individuals would have to be combined linearly. But the scale of an individual's utility function is arbitrary, so how can this be done? Others have taken my results and shown that this is the case even if all priors are the same. The likelihood function is public. That is all that can be agreed upon in a rational approach. The real problems of statistics, when to accept a hypothesis you know is false, is still very difficult. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists Wow! Herman is actually an economist who has done work on Arrow's Impossibility Theorem and the second welfarer theorem. I am more impressed! Is it true that statiistical decision theory has been largely discredited in the social sciences? >> If is meaningful to ask what is the probability that Bush will be >> re-elected in the 2008 election. A meaningful and operational >> value of p can be assigned to that future event, and it's likely >> that everyone will have a slightly different value. That is the >> crux of SUBJECTIVE probability. >That is entirely correct. But it all depends on the conditional: >IF it is meaningful to ask.... I don't think it IS meaningful, >except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot >be spoken of (IMHO) in a discussion between two or more people. >The utter failure of Bayesians to extend their methods to multi-person >or (better still) public conclusions, in spite of decades >of failed attempts, is highly significant. > There is an excellent reason; it cannot be done. There > is no method of arriving at a group decision if the > individuals and the group are both self-consistent other > than having one member decide. > Arrow proved this in his thesis, but did not include > randomization axioms. In my 1987 paper in _Statistics > and Decisions_, I pointed out that the problem to get > self-consistent group actions out of self-consistent > actions by the individual had to be Bayesian, in the > sense that the utility functions of the individuals > would have to be combined linearly. But the scale of > an individual's utility function is arbitrary, so how > can this be done? Others have taken my results and > shown that this is the case even if all priors are > the same. > The likelihood function is public. That is all that > can be agreed upon in a rational approach. The > real problems of statistics, when to accept a > hypothesis you know is false, is still very difficult. > -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: bayesians versus frequentists >Wow! Herman is actually an economist who has done work on Arrow's >Impossibility Theorem and the second welfarer theorem. I am more >impressed! I am not an economist, although I worked with economists, including Arrow. I was at the Cowles Commission for Research in Economics when I first established that a consistent choice scheme required Bayesian behavior. Later, I noticed the relation between consistent behavior with unknown state of nature given consistent behavior for each known state and the social welfare problem. >Is it true that statiistical decision theory has been largely >discredited in the social sciences? The social scientists, and many others, regard statistics as a set of procedures sent down on high, to be observed religiously. For many social scientists, the idea that there is no optimal program for mankind according to the values of the individuals is heresy. For many, even the idea that IQs need not be normally distributed is also heresy; most IQ scales do not reach 130 because the observations are converted using normality. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > since no such z exists, > therefore contradiction. > QED Ah, great. So there are also no solutions with n=2 or n=1. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: elementary proof of FLT x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > since no such z exists, > therefore contradiction. > QED > Ah, great. So there are also no solutions with n=2 or n=1. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Indeed, grobnerbases pushes the proof to the limit above and shows definitively that the only solutions have n = 0. -- c === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > since no such z exists, > therefore contradiction. > QED Since the equation x^n + y^n = z^n is assumed, for the context of Fermat's last theorem, to hold only at some particular value of z, not for all z or even all z in some open interval, it makes no sense to differentiate both sides of the equation with respect to z. To beat a dead horse here, z is not varying. There is no rate of change with respect to z. === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z it's perfectly ok to take differentials of integer equations, if both sides are related somehow, but it's not in this case... ln(x^n+y^n)=ln(z^n) taking differential of both sides w.r.t. z 0=n/z taking differential of both sides w.r.t. x n*x^n/(x*(x^n+y^n))=0 taking differential of both sides w.r.t. y n*y^n/(y*(x^n+y^n))=0 you can at best infer n=0 in this model, which contradicts nothing about x,y,z so no proof > since no such z exists, > therefore contradiction. > QED > Since the equation x^n + y^n = z^n is assumed, > for the context of Fermat's last theorem, to > hold only at some particular value of z, not for > all z or even all z in some open interval, it makes > no sense to differentiate both sides of the > equation with respect to z. > To beat a dead horse here, z is not varying. > There is no rate of change with respect to z. === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > it's perfectly ok to take differentials of integer equations, if both > sides > are related somehow, but it's not in this case... > ln(x^n+y^n)=ln(z^n) > taking differential of both sides w.r.t. z > 0=n/z > taking differential of both sides w.r.t. x > n*x^n/(x*(x^n+y^n))=0 > taking differential of both sides w.r.t. y > n*y^n/(y*(x^n+y^n))=0 > you can at best infer n=0 in this model, > which contradicts nothing about x,y,z > so no proof > since no such z exists, > therefore contradiction. > QED > Since the equation x^n + y^n = z^n is assumed, > for the context of Fermat's last theorem, to > hold only at some particular value of z, not for > all z or even all z in some open interval, it makes > no sense to differentiate both sides of the > equation with respect to z. > To beat a dead horse here, z is not varying. > There is no rate of change with respect to z. Love the handle, but I'm not sure what you are trying to defend. The fact that you were unable to obtain a contradiction by differentiating a fixed value with respect to what you imagine is a variable would not make the procedure mathematically tenable. How about a simple example: x = 1. Therefore 1 = 0 after differentiating both sides with respect to x. Or will you claim that in x = 1 the sides are not related somehow? This qualification (or disqualification, if you prefer) lacks clarity. If you want to differentiate an integer expression, you should first extend it to a function of real variables. Here the fact that x^n + y^n = z^n holds for some values of x,y,z,n promises nothing about an equation that holds for varying values of z (or any of the other unknowns). === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > it's perfectly ok to take differentials of integer equations, if both sides > are related somehow, but it's not in this case... > ln(x^n+y^n)=ln(z^n) > taking differential of both sides w.r.t. z > 0=n/z > taking differential of both sides w.r.t. x > n*x^n/(x*(x^n+y^n))=0 > taking differential of both sides w.r.t. y > n*y^n/(y*(x^n+y^n))=0 > you can at best infer n=0 in this model, > which contradicts nothing about x,y,z > so no proof > since no such z exists, > therefore contradiction. > QED > > Since the equation x^n + y^n = z^n is assumed, > for the context of Fermat's last theorem, to > hold only at some particular value of z, not for > all z or even all z in some open interval, it makes > no sense to differentiate both sides of the > equation with respect to z. > > To beat a dead horse here, z is not varying. > There is no rate of change with respect to z. > Love the handle, but I'm not sure what you are > trying to defend. The fact that you were unable > to obtain a contradiction by differentiating a > fixed value with respect to what you imagine > is a variable would not make the procedure > mathematically tenable. > How about a simple example: x = 1. > Therefore 1 = 0 after differentiating both sides > with respect to x. You can only take deriviatives if the equation f(x) = g(x) is true for all x in some interval. Someone else tried this last week in another thread, related to another problem. --- Christopher Heckman === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > > it's perfectly ok to take differentials of integer equations, if both sides > are related somehow, but it's not in this case... > ln(x^n+y^n)=ln(z^n) > taking differential of both sides w.r.t. z > 0=n/z > taking differential of both sides w.r.t. x > n*x^n/(x*(x^n+y^n))=0 > taking differential of both sides w.r.t. y > n*y^n/(y*(x^n+y^n))=0 > you can at best infer n=0 in this model, > which contradicts nothing about x,y,z > so no proof > > since no such z exists, > therefore contradiction. > QED > > Since the equation x^n + y^n = z^n is assumed, > for the context of Fermat's last theorem, to > hold only at some particular value of z, not for > all z or even all z in some open interval, it makes > no sense to differentiate both sides of the > equation with respect to z. > > To beat a dead horse here, z is not varying. > There is no rate of change with respect to z. > Love the handle, but I'm not sure what you are > trying to defend. The fact that you were unable > to obtain a contradiction by differentiating a > fixed value with respect to what you imagine > is a variable would not make the procedure > mathematically tenable. > How about a simple example: x = 1. > Therefore 1 = 0 after differentiating both sides > with respect to x. > You can only take deriviatives if the equation f(x) = g(x) is true > for all x in some interval. Someone else tried this last week in > another thread, related to another problem. > --- Christopher Heckman Yeah. Like I pointed out earlier in this thread: > Since the equation x^n + y^n = z^n is assumed, > for the context of Fermat's last theorem, to > hold only at some particular value of z, not for > all z or even all z in some open interval, it makes > no sense to differentiate both sides of the > equation with respect to z. grobnerbases seems (above) to defend taking differentiating both sides of an equation on the grounds that he failed to get any contradiction. I was merely demonstrating that one can. -- c === Subject: Re: elementary proof of FLT > I can't find anything wrong with this proof. > > proof by contradiction: > x^n+y^n=z^n (x,y,z,n the usual FLT conditions) > taking natural log of both sides > ln(x^n+y^n)=ln(x^n) > taking differential of both sides w.r.t. z > 0=1/z > > it's perfectly ok to take differentials of integer equations, if both sides > are related somehow, but it's not in this case... > ln(x^n+y^n)=ln(z^n) > taking differential of both sides w.r.t. z > 0=n/z > taking differential of both sides w.r.t. x > n*x^n/(x*(x^n+y^n))=0 > taking differential of both sides w.r.t. y > n*y^n/(y*(x^n+y^n))=0 > you can at best infer n=0 in this model, > which contradicts nothing about x,y,z > so no proof > > since no such z exists, > therefore contradiction. > QED > > Since the equation x^n + y^n = z^n is assumed, > for the context of Fermat's last theorem, to > hold only at some particular value of z, not for > all z or even all z in some open interval, it makes > no sense to differentiate both sides of the > equation with respect to z. > > To beat a dead horse here, z is not varying. > There is no rate of change with respect to z. > > Love the handle, but I'm not sure what you are > trying to defend. The fact that you were unable > to obtain a contradiction by differentiating a > fixed value with respect to what you imagine > is a variable would not make the procedure > mathematically tenable. > > How about a simple example: x = 1. > > Therefore 1 = 0 after differentiating both sides > with respect to x. > You can only take deriviatives if the equation f(x) = g(x) is true > for all x in some interval. Someone else tried this last week in > another thread, related to another problem. > Yeah. Like I pointed out earlier in this thread: Oops. I missed that. Sorry. --- Christopher Heckman === Subject: can someone help me out with this problem A $10,000 serial bond is to be redeemed in $1000 installments of principal per half year over the next 5 years. Interest at the annual rate of 12% is paid semiannually on the balance outstanding. How much should an investor pay for the bond in order to produce a yield rate of 8% convertibly semiannually Please help me out ASAP === Subject: Re: can someone help me out with this problem > A $10,000 serial bond is to be redeemed in $1000 installments of > principal per half year over the next 5 years. Interest at the annual > rate of 12% is paid semiannually on the balance outstanding. How much > should an investor pay for the bond in order to produce a yield rate of > 8% convertibly semiannually > Please help me out ASAP OK: I will help with some good advice. Read your textbook, and look over your lecture notes, to find similar problems that have already been solved. Talk to the TA, or to the course instructor. Go to the library. Sit down with a pad of paper and a pencil, write down what you already know about such problems, and see how this information will help you deal with the new case. Then tell yourself I will do this problem myself, because I realize that is the only way I will learn. R.G. Vickson Adjunct Professor, University of Waterloo === Subject: Re: perelman poincare proof looks really solid So now is it o.k. to copy/modify other people's preprints, and submit/publish it to some unknown journal ? In Chinese standard ? === Subject: Re: perelman poincare proof looks really solid >So now is it o.k. to copy/modify other people's preprints, and >submit/publish it to some unknown journal ? Hmm, are you saying that they copied/modified other people's preprint and submit/publish it to some unknown journal? I haven't read their paper and I would like to know exactly what wrongs they have done. > In Chinese standard ? What is Chinese standard by the way? -- Al Mok -- Aloysius K. Mok mok@cs.utexas.edu Department of Computer Science University of Texas at Austin Why? Why not? Austin, Texas 78712 (512)471-9542 === Subject: Re: perelman poincare proof looks really solid It is known that Perelman's work is not a complete proof. It is agreed by the experts that his work seems to be correct, though some details still needed to be worked out. The two Chinese have finally worked those details out. Their work has been checked by five Harvard's mathematicians, including S T Yau, a Field medalist. The title of their work is A Complete Proof of the Poincar.8e and Geometrization Conjectures - application of the Hamilton-Perelman theory of the Ricci flow http://www.intlpress.com/AJM/AJM-v10.php I think they have given proper credit to Perelman. For your interest, I am a Japanese, not Chinese. === Subject: Re: perelman poincare proof looks really solid It's a shame for Chinese scholars. === Subject: Re: Calculus XOR Probability <44732f25$0$24324$88260bb3@free.teranews.com> <447352c9$0$24287$88260bb3@free.teranews.com> <44748ee9$0$24320$88260bb3@free.teranews.com> unit infinity to be a suitable member of a number system, for where > its application _is_ both intuitive and analytically perfect. For > example, there was regular usage of such a notion for several thousands > of years. Really? Tell us where the notion that oo+oo=2oo, or oo+1>oo, was in regular use within the last few thousand years. === Subject: Re: Calculus XOR Probability Imaginatorium said: >> There is no difficulty whatsoever _naming_ the finite >> number which is the number of elements in N. I hereby name it Humphrey. >> What matters, however, is not Humphrey's name, but the fact that >> Humphrey does not exist, a fact that follows from an argument so >> simple, short, and oft-repeated, that anyone who really still cannot >> grasp it leaves one with little option than to assume said person >> (whoever they be) is really at least a little dim. > It's not a matter of grasping it in the sense of being able to follow the > logic, but there is a matter of measure that the argument ignores. If all > values are finite, then all differences are finite, and the number of finite > naturals from any one to any other is finite. There's no infinite sequence > there, quantitatively. Because there is no largest finite, because there is no > least upper bound to the set, the Dedekind definition for an infinite set > works, but because the set is restricted to finite values, with a constant > finite difference between successive values, the set cannot contain an infinite > NUMBER of unique finite naturals. > Maybe you think I'm dim, but you all appear > dim to me when you cannot grasp the simple notion that, if there is no element > an infinite number of positions beyond any other, then there is no infinite > number of elements in the set. It amazes us that you cannot grasp the blindingly simple logic behind the notion that given any n, n+1 exists, therefore there is no end to the n's, hence there are an infinitude of them. You have yet to offer any substantial refutation of this. If you want to make any headway in continuing with your beliefs, it would help immensely if you put your argument of infinite sets must have infinite distances between members into a formal argument. As it is, your English prose description is not logically sound. You'll find that to cast it into more formal language requires you to spell out your assumptions in much more precise detail, and that exercise is usually sufficient for most people to realize their mistakes. === Subject: Re: Calculus XOR Probability > There is no difficulty whatsoever _naming_ the finite >> number which is the number of elements in N. I hereby name it Humphrey. >> What matters, however, is not Humphrey's name, but the fact that >> Humphrey does not exist, a fact that follows from an argument so >> simple, short, and oft-repeated, that anyone who really still cannot >> grasp it leaves one with little option than to assume said person >> (whoever they be) is really at least a little dim. > It's not a matter of grasping it in the sense of being able to follow the > logic, but there is a matter of measure that the argument ignores. If all > values are finite, then all differences are finite, and the number of finite > naturals from any one to any other is finite. There's no infinite sequence > there, quantitatively. Because there is no largest finite, because there is no > least upper bound to the set, the Dedekind definition for an infinite set > works, but because the set is restricted to finite values, with a constant > finite difference between successive values, the set cannot contain an infinite > NUMBER of unique finite naturals. > Maybe you think I'm dim, but you all appear > dim to me when you cannot grasp the simple notion that, if there is no element > an infinite number of positions beyond any other, then there is no infinite > number of elements in the set. > It amazes us that you cannot grasp the blindingly simple logic behind > the notion that given any n, n+1 exists, therefore there is no end to > the n's, ... Oh, but I think that he can, at least up to this point. The problem is actually one of terminology, or rather of terminologology - that is, the metaproblem of what it means to say that a term means some particular thing. Mathematics is all about abstraction, so provided we define the terms we are using, we can discuss just about any coherent notion. Of course badly chosen terminology makes things hard to read, but choosing the wrong word simply cannot change the force of an argument. It's reasonably clear that somehow Tony does not grok this notion. Amusingly, Tony and Han agree with each other that they disagree totally about the existence of infinity, but I suspect that _actually_ they agree: it's just that since they have chosen different sets of personal terminology, they can't speak to each other about it. Roughly speaking, both of them are considering arithemetic on a finite binary register (i.e. a register with a leftmost bit and a rightmost bit), in some sort of limiting case as the register is made indefinitely long. I think it would be possible to formalise this carefully. Of course this is not what we mean by infinity, but it is not possible to say this to Tony, because he reinterprets the i-word into his own hazy definitions. (I have the impression that recently it's sort of dawned on him that his attempts at definition so far are circular. Perhaps we can hope...) Incidentally, you posted (approximately) the following theorem, recently. How does this fit with the fact that all numbers are red? [quoting DT] Theorem. The number of red naturals, |N|, is green. Proof. Assume that |N| is green. Then there exists a red natural m such that every member b in N is less than m, so b < m for all b in N. Since m is a red natural, it is a member of N. Since m is a member of N, it is either equal to or less than the largest member g of N, so m <= g. So there is a member g in N such that g >= m, but b < m for all b in N, which is a contradiction. So our assumption is false, and |N| is not red, so it is green. QED. Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability It amazes us that you cannot grasp the blindingly simple logic behind >> the notion that given any n, n+1 exists, therefore there is no end to >> the n's, ... > Oh, but I think that he can, at least up to this point. The problem is > actually one of terminology, or rather of terminologology - that is, > the metaproblem of what it means to say that a term means some > particular thing. Mathematics is all about abstraction, so provided we > define the terms we are using, we can discuss just about any coherent > notion. Of course badly chosen terminology makes things hard to read, > but choosing the wrong word simply cannot change the force of an > argument. It's reasonably clear that somehow Tony does not grok this > notion. [...] Yes, that seems to be a big part of Tony's problem. > Incidentally, you posted (approximately) the following theorem, > recently. How does this fit with the fact that all numbers are red? > [quoting DT] > Theorem. > The number of red naturals, |N|, is green. > Proof. > Assume that |N| is green. Then there exists a red natural m > such that every member b in N is less than m, so b < m for all > b in N. Since m is a red natural, it is a member of N. Since m > is a member of N, it is either equal to or less than the largest > member g of N, so m <= g. So there is a member g in N such > that g >= m, but b < m for all b in N, which is a contradiction. > So our assumption is false, and |N| is not red, so it is green. > QED. I think you misquoted me slightly. It should be: Theorem. The number of red naturals, |N|, is green. Proof. Assume that |N| is red. Then there exists a red natural m such that every member b in N is less than m, so b < m for all b in N. Since m is a red natural, it is a member of N. Since m is a member of N, it is either equal to or less than the largest member g of N, so m <= g. So there is a member g in N such that g >= m, but b < m for all b in N, which is a contradiction. So our assumption is false, and |N| is not red, so it is green. QED. This clearly shows that all naturals in N are red, but that |N| is green. Implicit in this proof is the definition: for natural n, red(n) iff there exists natural m such that red(m) and n < m. which is, as Tony is fond of saying, intuitively obvious. === Subject: Applications of integrative cognition As a college student at University of Virginia, I worked at a restaurant called the Tree House. All other employees were local black people, and I noticed one thing especially: They were all referring to each other by nickname. I asked why they did that, and the answers were such as What do you mean, why? and That's just how it is dog. One woman with intelligent eyes said, That's a good question. And indeed it is a good question to ask, one that is not limited to the Tree House. The statements like that's just how it is, dog reveal a common problem that is had by people who have lived in a single society all their lives: Lack of objective insight into its workings. Clearly referring to everybody by nickname is not the case in Long Island or in Sudan or in China. That's how it is, dog? Not exactly. Similar is the case with referring to things as normal or referring to one or another clique as speaking for everyone. What's normal in one place (calling people by nickname in the Tree House) is not what's normal in another (who does that in CIA?) And of course there is nobody who speaks for everyone (6 billion people), most of whom do not believe in anything similar to what is believed by the bully clique under consideration - and whose claims to speak for everyone are the heights of arrogance - the arrogance that they claim to denounce. The arrogance that it claims to be fighting but which it shows to a far more complete and far more overbearing extent, with less to justify it and far more horrid results for the people ensnared inside of it. The results being, that people can't leave the ghetto because they feel that would be selfish or arrogant; that people can't go into business or long-term work because they feel that would be collaborating with the white man; that people can't study science or engineering or teaching or literature because they feel that would be acting white; and what's left? Thuggery. Hideousness. Entrapment. All a necessary and inevitable result of the arrogance expropriated from individual to serve the tribal consciousness, and whose only possible outcome is war. I need hardly speak that dynamics similar to these, but in different forms, take place in different cultures besides the ghetto. Alexis De Tocqueville, a Frenchman who lived in America, came up with a balanced and highly insightful work into the nature of the United States. He did that because, as both an outsider and an insider, he was able to combine the external and the internal and thus to understand to a far more complete extent than is afforded by either mode of perception alone. As both outsider and insider, he was not subject to the error that is functional of either: The error of insider of failing to see the nation in its external effects and its workings - and the error of outsider of failing to experience it from within and thus failing to understand it. He was able to combine the objective and the subjective - to combine the experience and the analysis - to combine mind and heart, the two functionalities in which people discern and experience human reality. Both the subjective and the objective modes of understanding give part of the picture. But both are capable of error. The error inherent in first method is that of failure to quantify external effects - with the result that the people practicing the merely subjective perspective are constantly, and rightfully, accused of selfishness and lack of consideration for others. The error inherent in the second method is that of absolute lack of compassion for what is being studied; a lack of emotional closeness; a coldness and hideousness that is almost unhuman and that, lacking compassion by nature, therefore seeks to conquer and to subdue. This results in wars, dominations and extremecruelties and abuses. I am of the belief that complete understanding is that of understanding things in both in their external effects and their inner workings; a perspective possessing both objectivity and subjectivity - both intelligence and compassion - both mind and heart - both discerning it from without and experiencing it from within. The error of merely objective is twofold. First is failure to experience a phenomenon, resulting in lack of emotional insight. When people say they want someone to understand them, what they mean is that they want someone to feel how they feel rather than simply analyzing it, and certainly not analyzing them in reference to an external value set or a normative function or a psychological theory. When cultural and social and political entities say that they don't want outsiders telling them what to do, what they mean is that they want to first be understood in their experience before any agenda can be found that would be workable for them. But to do that and merely that is not adequate; external effects have to be quantified likewise. In combining objective observation and subjective experience, and then in interpolating between them, is arrived a complete picture of culture, religion, or social locale. Doing that, I refer to as integrative cognition. EXISTENTIAL BASIS There are sayings such as The only absolute is that there are no absolutes and all generalizations are false, including this one. This shows shortcomings in existentialism and relativism; however it does not render it worthless. The relative quality of reference frames - the curvature of the convex mirrors - do not make them worthless; it renders them incomplete. Which incompleteness is reduced through introduction of other reference frames that pose as check on the incompleteness. Through cross-referencing it becomes possible to see where each is right and where each is wrong. The convexity of the mirrors is quantified and reduced, and clarity rises. At which point becomes possible greater objectivity. And then it becomes possible to go to the next step: Look at the reasons for the convexity of the mirrors. To see the basic interests and ideologies functional thereto, and based on seeing these interests and ideologies for their character to see the extent to which the reflections are a function of relativistic distortion - distortion based on interests and ideologies underlying the mirrors - and to what extent these reflections are a function of clarity. Which interests and ideologies, being seen for what they are, are thus computed for the convexity that they produce; and the beams that make mirrors convex are gradually abraded until they become less prevalent, and greater objectivity is achieved. I am of the belief that this way the relative frames become closer to the absolute, and by tapping into Einsteinian relativity while understanding the nature thereof comes ever-greater clarity. By seeing the sun reflecting in multiplicity of convex mirrors, develops a closer and closer approximation of what is the sun. Convexities become visible; are understood for their essence; and are reduced. And the result is a more complete picture of the sun, and improvement in the quality of the mirrors. INTRINSIC AND REFERENTIAL There is the observation referential and there is the observation intrinsic. To know something from without, is to observe it referentially - referentially to a mode of experience, a mode of beliefs, a mode of cognition or a perceptual and conceptual base forming the human perception. To know it intrinsically, is to experience it on one's own heart or one's own skin. The integrative perspective is a synthesis of the referential and the intrinsic, combining assaying of the essential from multiple frames of reference with going into it intrinsically. Arriving at an assaying of a culture, a religion, a civilization or a mode of experience combining awareness of its experience with awareness of its external effects. This creates a complete representation of the phenomenon. One that is not limited by either failure to understand from within nor failure to understand from without, but rather creating a complete pictographical representation integrating the intrinsic and the referential. I shy from the use of the word objective, because I do not see complete objectivity as possible. Complete objectivity is a matter of being completely clear of all kinds of bias or agenda, and while that is a worthwhile pursuit I do not see it as being available for everyone, and certainly not something I've seen employed by anyone at all in either psychology or sociology. The reality is what I can touch and feel argument results in a perspective centered in self; which is not worthless but not complete, given the difference among selves. The reality is what can be reproduced argument results in perspective centered in canon; which is not worthless but not complete, given the change in lab conditions and explanations afforded by scientific progress. The reality is what refuses to go away when you stop believing in it argument fails to compute the extent to which people's beliefs shape their minds and thereafter their actions. This, once again, is not worthless but not complete. The solution I seek is one of approximation among reference frames. Not the nihilistic-anarchistic one of nothing-is-completely-objective-so-let's-burn-down-everything; but one rather of interpolation and check-and-balance among frames of reference, arriving at clarity of observation - combined with intrinsic, experiential understanding arrived at through direct experience of the mindset, the culture or the civilization that is being studied. A phenomenon can be assayed in reference to a method; it can be assayed in reference to an ideology; it can be assayed in reference to an interest; it can be assayed in reference to a a mode of thought and in reference to one's own experience. Each perspective forms a reference frame, akin to the Galilean reference frames of Einsteinian relativity. Each is a function of devices that shape it; each therefore creates a mirror in which is glimpsed the object of study. Each is valuable but not complete reality. Each, as such, is usable but not to the absoluteness. And enhancement of the external perspective is rendered through interpolation among these frames of reference, through checks and balances acting upon each other to attain at ever-greater clarity. There are of course many issues that come to pass. One is, Of what is one's own experience of the world a function? Is it not a function of previous experiences, assumptions, beliefs, characteristics, perspectives? Can it be ever truly objective? And is not greater objectivity arrived at through shifts of perspective, both mind and heart, as a result of interpolation, abrasion and ultimate integration among multiple modes of experience? What is the definition of integrity? It is that of acting as a single unit. A single unit premised on what? What premises are the correct ones, and will not something based on false premises come crashing down? The country I come from did. The static integrity of a rock can perhaps be defined as integrity, until one starts wondering what is under the rock and dislodges it, and then it crashes. I see integrity that is intellectually viable and far more interesting as one that is a dynamic synthesis; an ongoing whirling through chaos to create patterns to give to various interests; an integration and reconciliation leading toward meaningful outcomes and enhancement in people's experience of life. It is my belief that through continuing shifts of perspective and cross-examination between one's own experiences and the condition of many cultures, an interplay takes place between self and the object of study that refines one's experiential self toward greater clarity. Thus not only is study served, but also clarity of the self. The optimal result is exchange of ideas and understanding that helps to improve and to enrich both. Dangers include those of losing one's good qualities along with the bad ones if they are not part of the value of the culture at hand. Which qualities can be regained in a still greater and more informed manner when more informed, integrative, conclusions are drawn. The same applies likewise in philosophical inquiry. The existential theory of multiple truths - based on Einsteinian relativity - suggests this: That for greater reality it is important to experience and to assay all the truths in reference to each other, creating a greater understanding. And the theory of multiple Galilean reference forms, coming from physics, each frame a partial but incomplete incorporation of timespace, means that to know timespace to greater extent means to experience these forms and from reference point of these forms to draw connections and understanding of one another. If it is impossible to look at the sun for a prolonged period of time without going blind, then to see sun's reflections everywhere is a path toward greater understanding of the sun. And if the divine is invisible, then to see the reflections of the divine in the workings of cosmos as contained in everything from neutron stars to the plants to the oceans to human mind and human heart and inspired artwork is a path toward greater understanding and appreciation of the divine. The same applies to psychological inquiry. This is where R.D. Laing comes in handy: He believed in actually going into the person's inner world and seeing it from within. Indeed he showed several examples of how this could be done. For this, he is loved universally by the patients. The reason for that love? That he made an effort to understand them from within. Whereas, for example, Freud is universally hated - because he made no such effort. Instead, he partly observed, then he judged, then he stomped. Claiming himself a scientist, he performed grievous leaps of logic such as the following: Of leaping from analysis to judgment - of claiming that certain feelings, because they were supposedly (and in my belief wrongly) based in an earlier ego state, are therefore to be adjudicated infantile. It could be asked, Do my feelings exist for the sake of your values? Do my feelings exist for the sake of your supposed (and here obviously fallacious) rationality? Do I exist for your sake and the sake of your value system? And while we're speaking of value systems, what is the moral virtue of yours, Mr. Cocaine Addict? The merely objective - more correct term is referential - perspective is therefore necessarily cruel, precisely because it is merely referential. It is merely a result of observation in reference to existing beliefs. It sees that which is studied as an object - an object to be dissected and analyzed. And then it becomes the logic of dissection. And the question to be asked of all people engaging in such task (and those benefiting from it), according to their own devices, is: Of what is the logic a manifestation? What is the mechanism that runs the logic? How can it be linear when DNA, clouds, oceans, galaxies, stars, and the human body are anything but linear but full of curves and spirals and galaxies and multiplicitous interlocking ecosystemic forms? And is it rational to then claim that a single method is superior to all else in the universe when (A) it did not create the universe; (B) it did not create human brain that runs it; (C) any honest examination of such things fills one's heart with wonder at majesty and magnificence of the Universe and knowledge of the multiplicity of forms of logic present within it; and (D) knowing such things, to claim one form of cognition to be superior to all else in the Universe is not only irrational but indeed preposterous? The merely objective logic leads by the logic contained in its own devices indeed to the worst of narcissism and the worst of illogic. It leads to this: A dry, cruel, destructive maliciousness. Maliciousness that, having disconnected itself from the source of sensation, requires the intuitive and the emotional to prey upon - which it seeks out and, despising, goes on to destroy and discard. The illogic we see in abominations such as behaviorism, logical positivism, 1990s feminism and materialist fundamentalism. This is not accidental. This is not a stereotype. This is a necessary result of the mechanisms involved. If something exists at a rate greater than chance, then there must be causes for it. The probability theory dictates that, without causal factors, things will occur at the same rate as chance. So that, when they occur at a greater rate than chance, it stands to reason that there must be causal factors. And the way to discover the causal factors is through examination of fundamental mechanisms. The worst error of which merely objective method is capable is that of rationalistic judgmentalism. It is one thing to assay things in a scientific manner; it is quite another to judge mental processes - or ultimately life - as irrational because it runs by a logic different from the one that the person uses to study it. Thus we get to such foolish conclusions as Taken as a whole, the universe is absurd, women are stupid and we are the only two sane people in the universe, and sometimes I wonder about you. What rationality? If it's a rational universe then there must be a rationale, so let's find it out! Absurd? Absurd in reference toward what standard of thought? And is it not wrong to say that the problem is with the universe rather than with the method that fails to grasp it wholly, or with its incompetent utilization? Thus we get all kinds of hideousness coming to us through psychology. When one sees man's primary motivation as one of adequacy, one creates an ensnaring tribal consciousness that cannot bear what man has done through history: Enhance human condition. No man is an adequate match for a tiger; but he creates better technology to outsmart the tiger. To keep alive the tribal consciousness based on the lie of adequacy it becomes necessary to weed out those who would find better technology. Logical result, inevitable according to the logic contained in this method? Handicapper General, return to the roots, Ellsworth Toohey and Appalachia for us all. When one sees man's primary motivation as one of adaptation within social reality, one essentially deifies whatever order exists at the time. What needs to happen for that to remain the case? Kill off all the damn pesky troublemakers. Shove down people's throats the same party line. Make sure nobody thinks for themselves. Deadening, dumbening, degenerating of the population. Until the false God that is social consciousness turns into totalitarianism and becomes a torturer and a killer. Destroying everything that is passionate; destroying everything that is original; robbing people of their vitality; and turning the place into an emotional, intellectual and spiritual wasteland, in which man that the adaptation pretends to serve simply cannot exist and the claims of serving man become tranparently false. This is why it is entirely requisite, for any kind of objectivity, to see the situation in its entirety, rather than having it constituted in one or another mindset. That everyone in the ghetto will tell you the same thing, does not mean it is true; it means it is functional to the perspective of the ghetto. Of what is the perspective of ghetto a function? Let's find that out. The more that is done, the greater the objectivity and the greater the insight. PREMISES A mind torn from its premises is a mind that is insane. A mind put on another set of premises is a mind that is reconstituted. A mind that is intermingling among the reference frames of premises is confused, passionate, ecstatic, throbbing with fire of life. According to Keats, The way to intelligence is by making one's mind a thoroughfare for all thoughts - in which experiential chaos patterns form and are crystallized and can be used for any area of human endeavor. In what does one place one's faith? What are its premises? Is premise reason? Then don't reason's own investigations into the higher mathematics, biology, astronomy and everything else show the existence of other forms of rationality in the universe - its fundamental nonlinearity, instead its intricacies and complexities Is premise work? Then does not the process of work (and what it takes to maintain it) lead to all kinds of unanticipated consequences, which then require greater exploration, which then create a more enhanced understanding of what is work and what it is that is doing it, expanding both the definition of work and the definition of usable and worthwhile abilities? Is premise morality? Then does not the process of striving for ethical living lead to all kinds of agonizing moral choices, which likewise expand undestanding of human existence, and which when carried out incomptently or incompletely raise all kinds of moral questions about the validity of the morality that oneself claims to practice? Which morality, in order to be maintained through methods consistent with principle, including honesty of examination, then lead to a greater understanding and greater empathy and greater insight into just what constitutes truly ethical choice? Is premise responsibility? Does not acting in a responsible manner presuppose understanding the world enough that one knows the full range of consequences of one's actions and thus acts in a manner that anticipates and takes responsibility for the results? And are not the people who most loudly shout about responsibility hostile to knowledge and what it takes to attain it, thus putting a lie to their claim of valuing the responsibility they claim to espouse? A convex mirror reflects the sun partially truly and partially not truly. And to remove the beam from one's eye without replacing it with a brick, is to see its reflections in pools - including ones that don't contain in them the sulphur and the chlorine of the interest-shaped, belief-shaped thought contained in one's own culture. I don't regard it right to do away with things simply because they aren't complete. I don't regard it right to do away with logic, with economics, with politics; I seek to show the ways in which these methods can go right and the ways in which they can go wrong. I believe that all of these things, like all human things, are capable of both good and bad, and it is for that reason that we have a government of checks and balances. I believe that checks and balances could be applied on the totality of human activity and human thinking through complementary forms of cognition, in order that the result be both empathic and wise; in order that the capacity for wrong in each form be checked by balance with its opposite; in order that this synthesis lead to quality and refinement; and in order that in this manner we have minimax people living in minimax conditions. The samsara is beautiful and worthy of being loved and cherished. You don't kill a cat for not being a dog; you appreciate her as a cat, and then she can be happy as a cat and you can be happy as a lark having the cat - for having done what is required of intelligence: To appreciate the cat; to clear mind's conceptions from dog-shaped ideas of what cat should be; and thus to be able to reap of the beauty the cat wants to impart and would to those who are willing to understand without rushing to judgment. And if one's view of nature is a result of propaganda done by the dogs, then the process of understanding and loving the cat requires suspension of disbelief - suspension of judgment from dog-shaped premises of thought - and going into the world of the cat and there learning the beauty of which cat is herself capable. This, is the intrinsic side of the dialectic. Empathy. Going into the other person's world. Feeling the other person's experience of the world. Being Julia. Or - as Alexis De Tocqueville showed - doing another culture and doing that with it. REFERENTIAL ASSAYAL The referential side of the dialectic consists of assaying external results. This is best done from multiplicity of perspectives. The logical implication of relativity is that the way to approximate objectivity is to interpolate between reference frames. Thus, one does not observe American ghetto only from Northwest DC and the belief structure of Northwest DC; one does it also from China and Africa and the Cherokee tribe. Arriving at a more complete understanding of the ghetto. To greater elucidate what I'm talking about, let's conceptualize how a mindset of Northwest DC views and values. The mindset of Northwest DC, as a result of the ideologies in it contained, views and values according to standards of prosperity and human rights - the two great values preached by the American mindset. The mindset of China, on the other hand, views and values according to standards of family cohesiveness and work - its values, preached once again by Chinese ideology. So to assay the ghetto from the perspective of both China and Northwest DC, is to understand it more completely than it would be understood from either perspective (each value-set, belief-set, ideology-set, mindset shaping the convex mirror) acting alone. Thus is achieved a more complete representation of the American ghetto; one that, being more complete, thus becomes capable of more thorough understanding - and, pursuant this more thorough understanding, more complete and more valuable solutions for people in it living. To understand a culture - a mindset - an ideology, it is necessary to see it in both internal and external effects. It is necessary to understand the experience of those inside it and the way in which it affects the rest of the world and how other perspectives might see it. The other perspectives, in turn, must be seen for interests and values that are in them contained, in order that objectivity can be achieved. It is necessary to combine observation from multiple convex mirrors, with understanding of the nature of the convex mirrors, with subjective experience - and then weave. Interpolate. Integrate. Coming up with new patterns combining the mind and the heart, or analysis and experience. Creating an integrative understanding of the culture, combining external observation and internal experience. That a mirror is convex does not make it worthless. It simply means that it, like just about anything else, is incomplete; and to approximate objectivity it becomes necessary to either clear it completely and to remove the curvature at the core, or to reduce the curvature by assaying it in reference to other mirrors. Complete objectivity may not be achievable given the state of human existence; but it is most certainly capable of being approximated. The closer to the infinity, the more precise the result. Similarly different people's subjective experience of the cultures is going to differ according to what shapes their subjective worlds. The heart is a convex mirror just like the mind; that, like the mind, can be trained into greater clarity. The process of interpolation through external observation and subjective experience refines both the heart and the mind and reduces the quantity of beam contained in one's eye. Refinement of both organs becomes the result, and a path toward greater observational clarity. The quantity of beam is diminished also in the society, philosophy, person or institution being studied. As it sees its reflections in multiple mirrors, it cannot escape the fact of wrong that exists in it as well as right. As convexity of each mirror is reduced, more clear picture evolves; and as pictures become clear, it is no longer possible to blame all things for convexity. More importantly, the object is seen from more sides, with all the good and the bad as reflected in reference frames being made visible. At which point external picture becomes more and more complete. This is of course beneficial especially for writers and journalists. To be able to feel an experience from within, as opposed to merely observe it from without, is to have a far more complete understanding of it than would be found in merely objective (i.e. referential) perspective acting alone. When the participants in the Burningman event invited the media to partake in the festivities, that is precisely what they had in mind: To have the journalist feel - indeed get - the experience before writing about it in their papers. An event is described far more precisely when it is felt intrinsically, rather than simply assayed referentially to one or another convex mirror. And to be able to truly describe and to know a cat, it becomes necessary to remove the blinders imposed by the dog logic and feel the cat's world on one's heart and her fur on one's arms. At which point the journalist or the writer has the ability to integrate; to interpolate; to weave a gossamer thread contained of heart and mind between the internal and the particular external perspective to which he is catering. He can explain the experience in terms of the values and attitudes and convictions of the people for whom he is writing, while also making it palatable through his emotional understanding to their hearts. Having experienced it himself, he can speak for the people about whom he is writing, to the people for whom he is writing. He can be both analytical and empathic; at which point he can both feel and understand - and convey his knowledge of - the experience, to a far greater extent than he could from either perspective acting alone. PSYCHIATRY I need not say to what extent this can be of use to psychologists. This is a matter likewise of being able to both understand the patient's experience of the world, and to also see the entirety of the situation. Thus is arrived a far more complete plan of treatment than one of either acting alone. Thus the R.D. Laing method - of going into the person's world - is superior to the Freudian method of shackling it by assaying it through his dogma and presenting it in reference thereto. When a young woman in a mental hospital sat naked in the cell rocking and would not talk to anyone, he came in, took off his clothes and started rocking next to her, at which point she started talking to him as well. Upon leaving the cell he told other psychiatrists, Did you not think to do that? While this example may be considered by some to be ripe with potential for abuse of power, I am of the belief that this is a useful technique. Another technique I've seen used has been that of assaying the same from without. A highly gifted psychiatric nurse, when confronted with a client who thought himself a camel, was pressed to find out how to administer to him medication. She thought, Well, how would camel get his medication? and she put it in a cup on the floor. The camel hobbled up and took the medicine. What she had done, was figure out the logic of the psychosis and act upon it. She went to where he had gone, and then found the way to lead him out of there. She used her heart to empathize with the camel and her mind to figure out what to do about it. This, is the optimal of psychiatry combining the mind and heart. A question that the psychiatrist faces in front of him is, How does one defuse a nuclear bomb or a black hole? It's not enough to tinker with it on the outside as that would do nothing; it's not enough to go in it on the inside as one would get lost. The inner worlds can be beautiful or horrid or both at once; the question at this point becomes, How to reach into those inner worlds without breaking them, while leading the person out of the rabbit hole and into the sunlight? The error of many psychiatrists is that of completely destroying the people's worlds and leaving them dried, dead and broken. That's not psychiatry, that's oppression. It leaves the person worse off than he was before. But through interpolation between the referential and the intrinsic it becomes possible to go into the mechanism, while holding on from without so that one understands where to go with it. The outside of the bomb is seen reflected in multiplicity of convex mirrors that are referential perspectives - ones that cannot, I repeat, cannot, be limited to a single interest or social locale, but rather have to combine multiplicity of perspectives from multiplicity of places and states of consciousness - in order that outside effects can be approximated most closely. The quantity of beam in the eye is reduced through interpolation among these perspectives. Thus nearness to objectivity and some awareness of subjective experience is achieved, and mind is seen - to an extent, the greater the better - both from within and from without. At which point it's possible to go into the mind and to lead it out, while making it possible for the inner world to exist. Thus psychiatry that is worth anything depends upon the study of multiple perspectives, from multiple cultures and multiple theories. Integration through interpolation among these perspectives builds skill and wisdom in practice; which, due to the fluid nature of life, must be constantly refined. The quantity of beam in the eye is reduced through the process of integration and inter-analysis among Galilean reference frames: Of their assaying each other in reference to each other - of each viewing the other from its perspective and in so doing correcting each other's incompletenesses. The frames are themselves assayed based on the interest and the premise and the mentality of which they are a function, and quantity of distortion contained in them as functional of these things is quantified and assayed and computed. The curvature of the convex mirrors is quantified through analysis of such things, and correction is made in the resulting outcome. The beams in the eyes are reduced through this process, and greater clarity of reference frames is accomplished. And then it can be said that the matter is seen from the outside with a degree of objectivity. The subjective perspective meanwhile is figuring out how he feels and why. The goal of this process is to actually experience the world of the patient. This requires leaving behind one's conceptions - right ones or wrong ones - and taking the dive. While there, it is necessary to do the repair job; and then, like a skilled Houdini, it is necessary to go to the light, come out, and lead the patient out likewise. POLITICS The 1970s Democrats practiced an inclusive ideology that included multiplicity of perspectives. Jimmy Carter was accused of trying to micromanage, which is to say that he did not trust his subordinates, because he felt that they did not have the same perspective or agenda as did he. Ronald Reagan, on the other hand, felt comfortable delegating power to his subordinates, because he picked subordinates who believed exactly as he believed. This led to a well-run machinery; it also led to groupthink, coercion and similitude of all thought and attitude. Until it could be said, paraphrasing one of Reagan's subordinates, that If you've seen one Reagan Republican, you've seen them all. What are the dangers of groupthink? That of dogmatic thinking that fails to have clarity. That of prevailing on people into perpetuation of party line, thus blinding them to reality - reality that they claim to espouse and which they claim to be common sense. The party line effectively becomes commonsense, and any deviation from it is viciously prosecuted. Until anyone who has any capacity for original thinking is broken and desecrated and shoved into the gutter. The party line blinds one to reality; this perpetuates errors. All is assayed in reference to the party line rather than either intrinsically or through a synthesis of referential planes. At which point the task stops being one of leading and becomes one of bludgeoning. The population becomes devitalized; broken; expropriated. And all the perceptions, ideas, art, business, and thought of the civilization become functional to the beam under the convex mirror of Reaganism. Where does this lead? As in case of all ideologies that blind people to clarity, it led to consequences that were unanticipated (or sometimes perhaps anticipated) functional to the ideology. An ideology that believes that tax cuts are the solution to everything can, and always will, lead to huge federal debt. An ideology that believes that human activity cannot affect the environment can, and always will, lead to pollution and sickening of people here and elsewhere around the world. An ideology that believes that there is a single American way that is right for everyone can, and always will, lead to destruction of people's individuality and - far more ominously - people being prevented from contributing what they have to offer or making good on the Constitutional promise of life and liberty. These aren't coincidences or stereotypes. These are direct and inevitable results of these dogmas, and it is the Republicans therefore that need a reality check. So what do I prescribe? At political level another set of beam-removal techniques and interpolation. Synthesis through abrasion of beam. Attainment of clarity through this process and clarity that leads to enhancement and life and liberty functional to the noumenal, which is discerned likewise from within. And which therefore blossoms through the core. Ilya Shambatl === Subject: Re: Applications of integrative cognition A bunch of Polish scientists decided to flee their repressive government by >> hijacking an airliner and forcing the pilot to fly them to a western [...] >> Have patience. I'm just a simple pole in a complex plane. >> The plane later crashed because all the poles sat in the right half-plane. > Are you positive? Is this joke for real? === Subject: Re: Anyone got any good (or bad) mathematical jokes? >>A bunch of Polish scientists decided to flee their repressive government by >> hijacking an airliner and forcing the pilot to fly them to a western [...] >> Have patience. I'm just a simple pole in a complex plane. >> The plane later crashed because all the poles sat in the right half-plane. Are you positive? > Is this joke for real? Do you see any argument? Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: Anyone got any good (or bad) mathematical jokes? <873begpseb.fsf@nonospaz.fatphil.org> <87y7w5jz48.fsf@nonospaz.fatphil.org>A bunch of Polish scientists decided to flee their repressive government by >> hijacking an airliner and forcing the pilot to fly them to a western [...] >> Have patience. I'm just a simple pole in a complex plane. >> The plane later crashed because all the poles sat in the right half-plane. >> Are you positive? >> Is this joke for real? > Do you see any argument? I can't differentiate it from any other one. === Subject: Re: Integrative cognition Stay killfiled, moron & definitely keep this hallucinogenic insanity off >> scj. >> Susan > Susan Cohen! What's up with you, squirrelly-girl? > The squirrels are all hanging around your house, where the nuts are. > We don't want you on SCR either, scumblot. > Go plagiarize somewhere else. OOOH, territorial! Received-SPF: None; receiver=nym.alias.net; client-ip=69.119.204.244; envelope-from=; helo=bigapple.yi.org === Subject: (No Subject) Comments: This message did not originate from the above address. It was remailed by two or more anonymous mail services. Send complaints of REAL abuse ONLY to remaileradmin at optonline dot n NOTE: Trolling and flame wars are not considered real abuse. Mail-To-News-Contact: postmaster@nym.alias.net -=- This message was sent via two or more anonymous remailing services. === Subject: Re: Online Mathematics Encyclopedias <4etr4dF1gf422U1@individual.net I couldn't find Object Ring or Surrogate Factoring. > We'll just have to wait. > All beginnings are difficult. > Bob Kolker I hope they also put a way to contact the editor so one can report browsing the contents of EOM, it seems some entries were modified by various authors. --Titus Received-SPF: None; receiver=nym.alias.net; client-ip=69.119.204.244; envelope-from=; helo=bigapple.yi.org === Subject: Golbal moderation Comments: This message did not originate from the above address. It was remailed by two or more anonymous mail services. Send complaints of REAL abuse ONLY to remaileradmin at optonline dot n NOTE: Trolling and flame wars are not considered real abuse. Mail-To-News-Contact: postmaster@nym.alias.net -=- This message was sent via two or more anonymous remailing services. === Subject: Re: Golbal moderation > -=- > This message was sent via two or more anonymous remailing services. ***************** That's as good a plan for global moderation as I've seen anywhere, Mr. Fish. === Subject: Greater wealth per unit of resources The best unknown argument for the continuation of capitalism is that, over the time, the amount of resources necessary for creation of wealth, per unit of wealth, has been steadily reducing. A car took less resource expenditure than a house; a computer takes less resource expenditure than a car; and a modern cell phone with all the amenities takes less resource expenditure than any of the preceding. Thus, while continuing capitalism will go on consuming resources, then if the present trends hold it will do so at a lesser pace per unit of wealth. All new wealth created will be less resource-intensive than the existing wealth. And the economic growth could continue without turning into environmental disaster. There is hope still for something even better: Reduction in total consumption of resources, as new and more efficient technologies replace the old polluting ones and deliver greater utility for the same amount of resources consumed. Thus, hybrid cars will replace gas-guzzlers and save not only money but long-term good of the biosphere and of mankind; laptops will replace desktops; etc. There is even hope that the strip mall will be made obsolete through computerized selling and office through telecommuting, but I would not for that hold my breath. In order for capitalism to survive, it therefore needs to continue on the path of resource optimization. It is only in this way that it will be possible for mankind to survive and succeed in the new millenium, and for business prosperity to last and to mean anything. The new technologies will have to be energy- and resource-efficient, polluting at a lesser pace than existing ones, involving less harm than existing ones, creating greater wealth per unit of resources than existing ones, and thus producing more while consuming less. Now that, is a definition of progress worthy of pursuing. Ilya Shambat. === Subject: Re: Greater wealth per unit of resources > The best unknown argument for the continuation of capitalism is that, > over the time, the amount of resources necessary for creation of > wealth, per unit of wealth, has been steadily reducing. The best arguement for capitalism has been and always will be just three words, consider the alternative. Michael Gordge === Subject: Re: Greater wealth per unit of resources > The best unknown argument for the continuation of capitalism is that, > over the time, the amount of resources necessary for creation of > wealth, per unit of wealth, has been steadily reducing. A car took less > resource expenditure than a house; a computer takes less resource > expenditure than a car; and a modern cell phone with all the amenities > takes less resource expenditure than any of the preceding. Thus, while > continuing capitalism will go on consuming resources, then if the > present trends hold it will do so at a lesser pace per unit of wealth. > All new wealth created will be less resource-intensive than the > existing wealth. And the economic growth could continue without turning > into environmental disaster. Capiltolism is the biggest crock of . the Bayesian stooges of home economics ever produced. Since the tax subsidies for cell phones per unit volume, cost more than a cost of a house plus the cost of a car plus the cost of a home computer. Since of all the modern versions where invented by Wal-Mart Quantum Morons, rather than intelligent specimens of the species. > There is hope still for something even better: Reduction in total > consumption of resources, as new and more efficient technologies > replace the old polluting ones and deliver greater utility for the same > amount of resources consumed. Thus, hybrid cars will replace > gas-guzzlers and save not only money but long-term good of the > biosphere and of mankind; laptops will replace desktops; etc. There is > even hope that the strip mall will be made obsolete through > computerized selling and office through telecommuting, but I would not > for that hold my breath. > In order for capitalism to survive, it therefore needs to continue on > the path of resource optimization. It is only in this way that it will > be possible for mankind to survive and succeed in the new millenium, > and for business prosperity to last and to mean anything. The new > technologies will have to be energy- and resource-efficient, polluting > at a lesser pace than existing ones, involving less harm than existing > ones, creating greater wealth per unit of resources than existing ones, > and thus producing more while consuming less. > Now that, is a definition of progress worthy of pursuing. > Ilya Shambat. === Subject: Re: Greater wealth per unit of resources > Capiltolism is the biggest crock of . the Bayesian stooges > of home economics ever produced. So you have never shopped for a bargain? yeah right. > Since the tax subsidies for cell phones per unit volume, So in reality its not capitalism that is your problem, the redistribution of stolen money is. Michael Gordge === Subject: Re: Greater wealth per unit of resources > Capiltolism is the biggest crock of . the Bayesian stooges > of home economics ever produced. > So you have never shopped for a bargain? yeah right. I shopped for bargains all the time. I don't shop for bargain computers from morons like IBM though. Since the last thing they ever did with a computer was sell one to GM to use as a truck. > Since the tax subsidies for cell phones per unit volume, > So in reality its not capitalism that is your problem, the > redistribution of stolen money is. > Michael Gordge === Subject: Greater Wealth for the Rich Per Unit Wage Slave If capitalism is to survive, if disparity of wealth is to continue to increase, if the ancien regime is to continue to cling to power, then capitalists will need to keep newsgroups posters from exposing two scams: 1. the GOP artificial job shortage scam. 2. the corp. media cause celebre du jour scam Bret Cahill Can't figger out out why we get paid all this money to never figger anything out. -- corp. media editor === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave > If capitalism is to survive, if disparity of wealth is to continue to > increase, if the ancien regime is to continue to cling to power, then > capitalists will need to keep newsgroups posters from exposing two > scams: > 1. the GOP artificial job shortage scam. Um, could you tell me about that? === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave > If capitalism is to survive, if disparity of wealth is to continue to > increase, if the ancien regime is to continue to cling to power, then > capitalists will need to keep newsgroups posters from exposing two > scams: > 1. the GOP artificial job shortage scam. > Um, could you tell me about that? Now that you twisted my arm . . . :- ) By definition, a job is free market free trade where you get paid to do something you wouldn't ordinarily do. There are other situations, a hobbiest or White House intern or chattel slave does not get paid for his activity but the issue here is a paying job. How can there be a genuine job shortage when there isn't enough people to do the work for free, as a hobby? Unless everyone is infinitely wealthy then a job shortage is pure nonsense on it's face. In reality there's always work that needs to be done in any society so the problem isn't any genuine job shortage but a break down in communications. The truth is self evident: free speech precedes each and every free trade including the all important 14 figure a year free trade of employment at will. The way to expose the artificial job shortage scam is simple, easy, cheap, obvious and effective: Fix the break down in communications. Just sue for free speech under the First and Fifth Amendments. You'll get results if you win and you'll get results if the judges flee their courtroom. After all, you must be one bad political hombre to frighten federal appellate judges en banc in the land of the free and the home of the brave. That's why the nat'l government of the U. S. is in such a quandry now. All these 20th Century institutions are based on an assumption that cannot survive the info age. The assumption that there's a job shortage. Bret Cahill Bret Cahill === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 09:49:39 -0700, Bret Cahill something you wouldn't ordinarily do. Unless you are a politician and the job is to commit crime. -- Those who make peaceful revolution impossible will make violent revolution inevitable. --Sun Tzu === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave <448c5424.14159656@news-server.houston.rr.com> Would you call a biologist an expert in his field if he didn't even know what is designated by the acronym DNA? How about chemist who couldn't define stoichiometry? A mathematician who couldn't prove the Pythagorean theorem? A physicist who couldn't name Newton's Second Law of Motion? Just about everyone in their respective fields would call them outright frauds or just plain stupid. What am I saying? Most educated people outside their field would think they were frauds. That's because those questions are so basic you can go to any college or university and 98% of the high schools in the U. S. and you know you'll get the correct answer in less than 20 seconds. There would certainly be no stonewalling. Now, let's leave the reputable science and math departments at every last college and university on the planet and head on over the outspoken free market scholars at the Chicago School, von Mises Inst., Hoover Inst., American Enterprise, Cato, etc. and ask them a question that is even more basic to their field: Does free speech precede each and every free trade? Even though the correct answer is an obvious self evident truth, the outspoken market economists won't have any answers. In fact, these scholars will stonewall and dodge like Enron executives before a Senate panel. Now, why oh why are we letting these two bit rinky dink third world flim flam swindlers play any role whatsoever in deciding the economic policy of a great nation? All it would take is a few phone calls and no one would ever hear from these GOP scammers again. Bret Cahill === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 09:49:39 -0700, Bret Cahill India There's a shortage of people in the US who will package medical devices for less than 2 dollars per hour. Job shortage - true Solution ---> Mexico There's a shortage of companies in the US who will provide plastic part tooling and/or the parts themselves for 1/2 the price US companies would charge. Resource shortage = true Solution ---> China Anyway. It's going to start raining here soon. Alberto. My fruit trees are getting psyched up. bob - there's a pineapple outside that may get harvested later today === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave India Markets level wealth, between nations as well as between the rich and poor in a nation, but since ave. mean income is well over $50/hour in the United States American employees need only focus on using markets to level wealth _in_ the U. S. to get all U. S. incomes much closer to $50/hour. This will also have two other beneficial effects for employees: 1. it will grow the economy overall so your income will be heading to a ave. mean much higher than $50/hr, and 2. politicians will no longer have any interest in using free trade and open borders to jerk the wages of employees down. If it's good for most Americans to close the borders and have trade restrictions then the politicians will do it. But nothing's gonna happen until you pull out the 9' loaded blacksnake whip aka the great charter of American freedom and apply it to the fannies of politicians so that markets work in the U. S.. Bret Cahill There's going to be a leveling between rich and poor, between rich and poor nations . . . [Don't, don't -- I know what you're thinking -- don't even think you can stop it.] -- IBM CEO Lou Gerstner before Congress (May 1999) in the most widely ignored statement ever === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 10:44:03 -0700, Bret Cahill whip aka the great charter of American freedom and apply it to the >fannies of politicians so that markets work in the U. S.. In East Texas there is another way to deal with corrupt politicians. That's why everyone carries a chain in the back of their pickup. -- Those who make peaceful revolution impossible will make violent revolution inevitable. --Sun Tzu === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 10:44:03 -0700, Bret Cahill <448c82c9.26100843@news-server.houston.rr.comsince ave. mean income is well over $50/hour > Mean income is a poor indication of the distribution of wealth True. It only shows average mean income. >because > the distribution of income is skewed. And people find out it is skewed when they see that ave. mean income is several times median income. > That's why the govt used median > income. Until I appeared the corp. whore media used median income to keep the people from finding out how much they were being ripped off. They don't try that scam anymore. Bret Cahill === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 20:56:47 -0700, Bret Cahill since ave. mean income is well over $50/hour >> Mean income is a poor indication of the distribution of wealth >True. It only shows average mean income. >>because >> the distribution of income is skewed. >And people find out it is skewed when they see that ave. mean income is >several times median income. >> That's why the govt used median >> income. >Until I appeared the corp. whore media used median income to keep the >people from finding out how much they were being ripped off. Bret Cahill to the rescue! >They don't try that scam anymore. Congratulations! === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave >On 11 Jun 2006 20:56:47 -0700, Bret Cahill >since ave. mean income is well over $50/hour > Mean income is a poor indication of the distribution of wealth >>True. It only shows average mean income. >because > the distribution of income is skewed. >>And people find out it is skewed when they see that ave. mean income is >>several times median income. > That's why the govt used median > income. >>Until I appeared the corp. whore media used median income to keep the >>people from finding out how much they were being ripped off. >Bret Cahill to the rescue! >>They don't try that scam anymore. >Congratulations! Wow, it's a been a long time since I've seen msg from Cahill. It appears the loony is back! -- The trouble with the world is that the stupid are cocksure and the intelligent are full of doubt -- Bertrand Russell === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave <448c82c9.26100843@news-server.houston.rr.com> since ave. mean income is well over $50/hour >> Mean income is a poor indication of the distribution of wealth >True. It only shows average mean income. It only shows what the average person would be making if he wasn't getting ripped off. >>because >> the distribution of income is skewed. >And people find out it is skewed when they see that ave. mean income is >several times median income. >> That's why the govt used median >> income. >Until I appeared the corp. whore media used median income to keep the >people from finding out how much they were being ripped off. > Bret Cahill to the rescue! My altruism is motivated purely by self interest: It's like the Democratic Party bumper sticker: The ass you save may be your own. >They don't try that scam anymore. > Congratulations! The true populist encourages everyone, even libtards to get involved politically but you'll need to read something smarter than Rand to be effective. Bret Cahill === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 12 Jun 2006 14:00:57 -0700, Bret Cahill >since ave. mean income is well over $50/hour > Mean income is a poor indication of the distribution of wealth >>True. It only shows average mean income. >It only shows what the average person would be making if he wasn't >getting ripped off. >because > the distribution of income is skewed. >>And people find out it is skewed when they see that ave. mean income is >>several times median income. > That's why the govt used median > income. >>Until I appeared the corp. whore media used median income to keep the >>people from finding out how much they were being ripped off. >> Bret Cahill to the rescue! >My altruism is motivated purely by self interest: >It's like the Democratic Party bumper sticker: >The ass you save may be your own. >>They don't try that scam anymore. >> Congratulations! >The true populist encourages everyone, even libtards to get involved >politically but you'll need to read something smarter than Rand to be >effective. Winnie the Pooh? === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 10:44:03 -0700, Bret Cahill Solution --> India >Markets level wealth, between nations as well as between the rich and >poor in a nation, but since ave. mean income is well over $50/hour in >the United States References? >American employees need only focus on using markets >to level wealth _in_ the U. S. to get all U. S. incomes much closer to >$50/hour. Heh >This will also have two other beneficial effects for employees: >1. it will grow the economy overall so your income will be heading to >a ave. mean much higher than $50/hr, and >2. politicians will no longer have any interest in using free trade >and open borders to jerk the wages of employees down. If it's good for >most Americans to close the borders and have trade restrictions then >the politicians will do it. >But nothing's gonna happen until you pull out the 9' loaded blacksnake >whip aka the great charter of American freedom and apply it to the >fannies of politicians so that markets work in the U. S.. You are entertaining. Please continue? === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave Solution --> India >Markets level wealth, between nations as well as between the rich and >poor in a nation, but since ave. mean income is well over $50/hour in >the United States > References? DeTocqueville (1833) and Gerstner (1999) on markets leveling wealth. As for ave. mean income per working hour take the GNP per year and divide by the total hours worked per year for a rough number. >American employees need only focus on using markets >to level wealth _in_ the U. S. to get all U. S. incomes much closer to >$50/hour. > Heh The high tax Clinton economic boom, the longest economic expansion in the history of the republic, made a lot of people happy. >This will also have two other beneficial effects for employees: >1. it will grow the economy overall so your income will be heading to >a ave. mean much higher than $50/hr, and >2. politicians will no longer have any interest in using free trade >and open borders to jerk the wages of employees down. If it's good for >most Americans to close the borders and have trade restrictions then >the politicians will do it. >But nothing's gonna happen until you pull out the 9' loaded blacksnake >whip aka the great charter of American freedom and apply it to the >fannies of politicians so that markets work in the U. S.. > You are entertaining. Please continue? Clinton approval: 70% (and increasing) AwOL Bush approval: 28% (and dropping) Care for some more? Bret Cahill === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave On 11 Jun 2006 11:36:31 -0700, Bret Cahill AwOL Bush approval: 28% (and dropping) If you believe those figures, you have lost all credibility. -- Those who make peaceful revolution impossible will make violent revolution inevitable. --Sun Tzu === Subject: Re: Greater Wealth for the Rich Per Unit Wage Slave >> If capitalism is to survive, if disparity of wealth is to continue to >> increase, if the ancien regime is to continue to cling to power, then >> capitalists will need to keep newsgroups posters from exposing two >> scams: >> 1. the GOP artificial job shortage scam. >Um, could you tell me about that? Clue 1: H1-B and L1 visas Clue 2: cross reference Clue 1 with Congress It's all a conspiracy! heh bob - but it's not interesting enough for a reality show === Subject: Re: Greater wealth per unit of resources > In order for capitalism to survive, it therefore needs to continue on > the path of resource optimization. For capitalism to survive? IMHO, capitalism is the null-hypothesis, it is the lack of social restrictions. Saying for capitalism to survive is like saying for vacuum to survive. === Subject: Re: Greater wealth per unit of resources http://www.factor10-institute.org/pdf/F10REPORT.pdf === Subject: Re: Greater wealth per unit of resources <2kcl829gd766jejugvahqr4te7s0t9t50d@4ax.com http://www.factor10-institute.org/pdf/F10REPORT.pdf 67 pages is all one needs to know. One word: Condom But hey, I'm a fan of Occam. -tg === Subject: Re: Greater wealth per unit of resources >> http://www.factor10-institute.org/pdf/F10REPORT.pdf >67 pages is all one needs to know. >One word: Condom >But hey, I'm a fan of Occam. Not of Trujillo. === Subject: Re: Greater wealth per unit of resources <2kcl829gd766jejugvahqr4te7s0t9t50d@4ax.com http://www.factor10-institute.org/pdf/F10REPORT.pdf > 67 pages is all one needs to know. > One word: Condom > But hey, I'm a fan of Occam. > -tg Count a consumer of a cooking oils per family and even singles. === Subject: Re: Greater wealth per unit of resources > http://www.factor10-institute.org/pdf/F10REPORT.pdf THX interesting report, what i read. cpl important comments were ... on page 12 if china undertook to have the same number of cars as America, they would need 20% of their *arable* land for roads and car park spaces. if the entire world of 6 billion aspired to being as *rich* as the consumers in the *advanced* nations, it would require 3 new planet earths in order to provide the necessary resources. the theory they suggest to change the situation makes a lot of rational sense imho. The worse things get, the greater the will to change current practices. time will tell. cheers. === Subject: Re: Greater wealth per unit of resources > The best unknown argument for the continuation of capitalism is that, > over the time, the amount of resources necessary for creation of > wealth, per unit of wealth, has been steadily reducing. A car took less > resource expenditure than a house; a computer takes less resource > expenditure than a car; and a modern cell phone with all the amenities > takes less resource expenditure than any of the preceding. This is idiotic. What does a car have to do with a house? A small house takes less resources than a big house. And yet, with the continuation of capitalism, we build bigger houses. Why don't you explain how this works? -tg > Thus, while > continuing capitalism will go on consuming resources, then if the > present trends hold it will do so at a lesser pace per unit of wealth. > All new wealth created will be less resource-intensive than the > existing wealth. And the economic growth could continue without turning > into environmental disaster. > There is hope still for something even better: Reduction in total > consumption of resources, as new and more efficient technologies > replace the old polluting ones and deliver greater utility for the same > amount of resources consumed. Thus, hybrid cars will replace > gas-guzzlers and save not only money but long-term good of the > biosphere and of mankind; laptops will replace desktops; etc. There is > even hope that the strip mall will be made obsolete through > computerized selling and office through telecommuting, but I would not > for that hold my breath. > In order for capitalism to survive, it therefore needs to continue on > the path of resource optimization. It is only in this way that it will > be possible for mankind to survive and succeed in the new millenium, > and for business prosperity to last and to mean anything. The new > technologies will have to be energy- and resource-efficient, polluting > at a lesser pace than existing ones, involving less harm than existing > ones, creating greater wealth per unit of resources than existing ones, > and thus producing more while consuming less. > Now that, is a definition of progress worthy of pursuing. > Ilya Shambat. === Subject: Re: Greater wealth per unit of resources Clearly if you increase disparity of wealth as well as the size of the economy most become poor and consume less resources. Several have already pointed out that W. Bush is really an Earth Firster. Bret Cahill === Subject: Re: Greater wealth per unit of resources >> The best unknown argument for the continuation of capitalism is that, >> over the time, the amount of resources necessary for creation of >> wealth, per unit of wealth, has been steadily reducing. A car took less >> resource expenditure than a house; a computer takes less resource >> expenditure than a car; and a modern cell phone with all the amenities >> takes less resource expenditure than any of the preceding. > This is idiotic. What does a car have to do with a house? > A small house takes less resources than a big house. And yet, with the > continuation of capitalism, we build bigger houses. Why don't you > explain how this works? (...)the amount of resources necessary for creation of wealth, per unit of wealth(...) -- Michaü Gancarski If there's not drama and negativity in my life, all my songs will be really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources The best unknown argument for the continuation of capitalism is that, >> over the time, the amount of resources necessary for creation of >> wealth, per unit of wealth, has been steadily reducing. A car took less >> resource expenditure than a house; a computer takes less resource >> expenditure than a car; and a modern cell phone with all the amenities >> takes less resource expenditure than any of the preceding. > This is idiotic. What does a car have to do with a house? > A small house takes less resources than a big house. And yet, with the > continuation of capitalism, we build bigger houses. Why don't you > explain how this works? > (...)the amount of resources necessary for creation of wealth, per unit > of wealth(...) Except for the trivial observation that the volume of a house increases in a non-linear fashion with the number of surface elements, what about *capitalism* makes the increased volume use less resources? A 2x6 is a 2x6. -tg -- > Michañ Gancarski > If there's not drama and negativity in my life, all my songs will be > really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources [...] >> This is idiotic. What does a car have to do with a house? >> A small house takes less resources than a big house. And yet, with the >> continuation of capitalism, we build bigger houses. Why don't you >> explain how this works? >> (...)the amount of resources necessary for creation of wealth, per unit >> of wealth(...) > Except for the trivial observation that the volume of a house increases > in a non-linear fashion with the number of surface elements, what about > *capitalism* makes the increased volume use less resources? A 2x6 is a > 2x6. (...)per unit of wealth(...) -- Michaü Gancarski If there's not drama and negativity in my life, all my songs will be really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources > This is idiotic. What does a car have to do with a house? >> A small house takes less resources than a big house. And yet, with the >> continuation of capitalism, we build bigger houses. Why don't you >> explain how this works? >> (...)the amount of resources necessary for creation of wealth, per unit >> of wealth(...) > Except for the trivial observation that the volume of a house increases > in a non-linear fashion with the number of surface elements, what about > *capitalism* makes the increased volume use less resources? A 2x6 is a > 2x6. > (...)per unit of wealth(...) You could save some bandwidth by being less cute. My unit of wealth is linear feet in the perimeter of my house. What's yours? -tg > -- > Michañ Gancarski > If there's not drama and negativity in my life, all my songs will be > really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources >> [...] > This is idiotic. What does a car have to do with a house? >> A small house takes less resources than a big house. And yet, with >> the > continuation of capitalism, we build bigger houses. Why don't you > explain how this works? >> (...)the amount of resources necessary for creation of wealth, per >> unit > of wealth(...) >> Except for the trivial observation that the volume of a house >> increases >> in a non-linear fashion with the number of surface elements, what >> about >> *capitalism* makes the increased volume use less resources? A 2x6 is a >> 2x6. >> (...)per unit of wealth(...) > You could save some bandwidth by being less cute. My unit of wealth is > linear feet in the perimeter of my house. What's yours? Resources needed to build (produce) a house are not only some 2x6s or 2x4s. It is the total amount of resources used in every stage of production of every single component of the house. Sure, if you need six 2x6s, you will probably need them anyway but improvements are easily tracable in tools used to build it or in efficiency of companies producing using a pen (let us assume that there are no other methods) you will probably use the same amount of ink per letter and there is not much you can do about this. But you can stiil organise your work better if you, for example, decide to write it during daytime (consuming less electric power) and/or to split the process in well planned phases. So yes, two persons writing similar texts can use exactly the same amount of ink but the ink is not the only resource needed. There are many more, including those used by ink factories and pen producers. And they, too, can and will improve their methods of production in pursue for a greater profit. -- Michaü Gancarski If there's not drama and negativity in my life, all my songs will be really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources [...] > This is idiotic. What does a car have to do with a house? >> A small house takes less resources than a big house. And yet, with >> the > continuation of capitalism, we build bigger houses. Why don't you > explain how this works? >> (...)the amount of resources necessary for creation of wealth, per >> unit > of wealth(...) >> Except for the trivial observation that the volume of a house >> increases >> in a non-linear fashion with the number of surface elements, what >> about >> *capitalism* makes the increased volume use less resources? A 2x6 is a >> 2x6. >> (...)per unit of wealth(...) > You could save some bandwidth by being less cute. My unit of wealth is > linear feet in the perimeter of my house. What's yours? > Resources needed to build (produce) a house are not only some 2x6s or > 2x4s. It is the total amount of resources used in every stage of > production of every single component of the house. Sure, if you need six > 2x6s, you will probably need them anyway but improvements are easily > tracable in tools used to build it or in efficiency of companies producing > materials required to build this house. In other words, your definition of unit of wealth and resources is whatever supports your statements but not whatever doesn't support your statements. If you build a bigger house, you use more resources (again, controlling for geometry). This is true under capitalism, socialism, feudalism, or whatever system you like. If you wish to make a case, you need to fix your definitions at the beginning and stick to them. My measure of wealth is the linear feet in the perimeter of my house. I consider resources to be physical things which are consumed in producing those linear feet of wall. (Not the cost of those things.) Please give those easily traceable to capitalism improvements that make the ratio of resources to wealth diminish. -tg > using a pen (let us assume that there are no other methods) you will > probably use the same amount of ink per letter and there is not much you > can do about this. But you can stiil organise your work better if you, for > example, decide to write it during daytime (consuming less electric power) > and/or to split the process in well planned phases. So yes, two persons > writing similar texts can use exactly the same amount of ink but the ink > is not the only resource needed. There are many more, including those used > by ink factories and pen producers. And they, too, can and will improve > their methods of production in pursue for a greater profit. > -- > Michañ Gancarski > If there's not drama and negativity in my life, all my songs will be > really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources [...] >> Resources needed to build (produce) a house are not only some 2x6s or >> 2x4s. It is the total amount of resources used in every stage of >> production of every single component of the house. Sure, if you need six >> 2x6s, you will probably need them anyway but improvements are easily >> tracable in tools used to build it or in efficiency of companies >> producing >> materials required to build this house. > In other words, your definition of unit of wealth and resources is > whatever supports your statements but not whatever doesn't support your > statements. In other words, nope. > If you build a bigger house, you use more resources (again, controlling > for geometry). This is true under capitalism, socialism, feudalism, or > whatever system you like. This is, obviously, not true. There are many ways to build a house (or to try tu build one) and one person can spend much more resources building nothing than some other one that will have the task acomplished. And the difference between a house and no house at all is much bigger than the difference between a house and a bigger house. > If you wish to make a case, you need to fix your definitions at the > beginning and stick to them. How should I fix your imaginations about my definitions? > My measure of wealth is the linear feet in > the perimeter of my house. I consider resources to be physical things > which are consumed in producing those linear feet of wall. (Not the > cost of those things.) But the cost of those things is the most universal and efficient way to determine the amount of resources used. You still try to forget that what really counts is every stage of production of every component of the house. > Please give those easily traceable to > capitalism improvements that make the ratio of resources to wealth > diminish. Try to imagine yourself producing one of those new Ford GTs. I can be almost sure that you would consume much more resources to produce this car than Ford factories do. Why? Try and guess it for yourself. The thing that is easily tracable to free market activity is growing efficiency of production which means less total resources used for one unit of total wealth produced. -- Michaü Gancarski If there's not drama and negativity in my life, all my songs will be really wack and boring or something. Eminem === Subject: Re: Greater wealth per unit of resources If your measurement of wealth is in dollars you can increase wealth by simply by printing money. Isn't that what Wall Street is worried that Bernanke is doing? Bret Cahill === Subject: Re: Greater wealth per unit of resources > The best unknown argument for the continuation of capitalism is that, > over the time, the amount of resources necessary for creation of > wealth, per unit of wealth, has been steadily reducing. A car took less > resource expenditure than a house; a computer takes less resource This development cannot be credited to capitalism but is a result of science and technology. In fact, capitalism with its property rights, copyright and patent laws, have hindered and slowed down the technical progress. An example is the steam driven pump developed by a dutchman in the earliest mechanized industry development, which could not be used until a hundred years after the inventor had died, because of problems with patents and property rights. Big companies buy up millions of patents and put them in a drawer, to stop their competitors from using them, property owners have more power than inventors which hinder the development, etc.. The first man who built and flew a motorized aeroplane was Gustave Whitehead, who flew his homebuilt aeroplanes years before the Wright brothers. http://en.wikipedia.org/wiki/Gustave_Whitehead Here is the story: His workshop assistant later described how the Wright brothers for the second time had visited Whitehead to discuss financing his aeroplane. When they left Whitehead said to his assistant: I bet those rascals will never finance my aeroplane anyway, and now I have told them all my secrets. Later the Wright brothers forced the Smithsonian institute to sign a contract forbidding the institute to ever mention any motorized flights before the Wright brothers first flight. The Wright brothers were more lawyers than inventors. They dug up all information they could get hold of from real inventors, and when they finally got an aeroplane into the air, a short jump before the plane crashed, they filed for patents and used their tricky business minds to make sure they would be remembered as the inventors of the aeroplane. Whitehead had flown his nr 21 aeroplane more than two years earlier. It was a safe and stable aeroplane which did not need a catapult or a strong headwind to start. It landed safely on land as well as on water. In August 1901 he flew 800 meter and landed on a free spot in Bridgeport suburbs, turned the plane around, started again and flew back to the place he started from. He did four flights that day, which later lead to conflicting statements from witnesses. One of the witnesses was a reporter from the newspaper The Bridgeport Herald, and he described one of the 800 meter flights in the newspaper. be read in the lower part of the web page referred to above. -- Roger J. === Subject: Re: Greater wealth per unit of resources > In fact, capitalism with its property rights, copyright and patent > laws, have hindered and slowed down the technical progress. You may be thinking of patent laws interpreted in a libertarian sense as in greedy U. S. drug companies who get continuation patents on the color of their pills. The original intent of the patent claims as outlined in Art. I, Sec. 8 of the U. S. Constitution was to promote the useful arts not to make inventors rich by crushing the innovation of other inventors. Everyone knew something was wrong with Microsoft but since the all truth comes from money mentality prevails in the U. S., the government wasn't able to formulate the proper constitutional remedy. Instead they tried to use anti-trust laws which are irrelevant. But properly interpreted patent laws definitely encourage innovation. Patents go back to the ancient Greeks. > An example is the steam driven pump developed by a dutchman in the > earliest mechanized industry development, which could not be used until > a hundred years after the inventor had died, because of problems with > patents and property rights. Any area concerning money will inevitably produce a lot of rip offs and injustices. The proper remedy is to figure out how to reduce the rip offs, not patent rights. > Big companies buy up millions of patents and put them in a drawer, to > stop their competitors from using them, Any examples of stuff that appeared only 14 years after the patent claim? Here's an example of how patents work: Back in the early 1960s a famous inventor, Dr. Land, patented a great invention: windshields or visors polarized 90 degrees off polarized headlights so that oncoming traffic with the same system would not be blinded by your headlights at night. Something like that, however, requires everyone to be on board so Land tried to get Dept. of Transportation to _require_ everyone to buy his patented invention where Land could charge anything he wanted. Land could shut down the auto business. DoT didn't go for that one. The patent has long expired so no one is interested in promoting an idea that could have saved thousand of lives. An inventor is not necessarily doing society a favor by giving away inventions. Even low tech stuff for the 3rd world should be patented. > property owners have more power > than inventors which hinder the development, etc.. > The first man who built and flew a motorized aeroplane was Gustave > Whitehead, who flew his homebuilt aeroplanes years before the Wright > brothers. > http://en.wikipedia.org/wiki/Gustave_Whitehead > Here is the story: > His workshop assistant later described how the Wright brothers for the > second time had visited Whitehead to discuss financing his aeroplane. > When they left Whitehead said to his assistant: > I bet those rascals will never finance my aeroplane anyway, and now I > have told them all my secrets. If you want to take credit for your inventions, you aren't supposed to be keeping secrets. You are supposed to show diligence in PUBLISHING your discoveries. That's the whole point of patenting. I post ideas all the time that I don't have time to persue -- if it's not 12 figures/year I cannot justify taking time off from politics -- but I never post enough information on how to actually build anything. I never seem to quite understand the technology. Why? Because it leaves the door open for someone else to patent and a product has a better chance of being developed if it has patent protection. Inventors feed off more than they compete against each other and they serve themselves as well as other inventors as well as the rest of the collective by encouraging patent protection. Now what could be more socialist than that? Bret Cahill === Subject: Re: Greater wealth per unit of resources It's called cross posting, a low class form of trashing. You always do this with real resource cheap low class wealth style. === Subject: Re: Greater wealth per unit of resources Are there solutions to the number-theoretic equation: >sigma(sigma(A^2)) = 2*(A^2), where A is odd? None with A < 150000. >Note that, if we don't restrict A to be odd, then A = 2^k is a solution >when sigma(A^2) is prime (i.e. a Mersenne prime). No even solutions < 100000 except those, either. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Verification needed, simple algebra > I've been interested in solving problems, making a LOT of mistakes > along the way. > That's just how it is, problem solving is more like Edison's search for > the light bulb than most people's idea of a brilliant insight in the > middle of the night. > You throw everything but the kitchen sink at problems and then try that > as well, if you're really serious, and I am. > But it makes it easy for other people to target you, ridicule you, and > convince others that you are a crackpot, loon or crank. The thing is edison didn't go public with his mistakes. he tested them first. > James Harris -- Bye. Jasen === Subject: Re: Looking for a surjection or R^2 What about f(x,y) = e^(iy)(e^x+1) + y? It's easy to check that the Jacobian is > 0 and f is not bijective: f(a,pi) = f(b,3pi) = 0 for some a, b. The sujectivity seems OK from some geometrical evidence: the map e^(iy)(e^x+1) winds the plane onto the complement of the unit circle, and adding y to it, will cover each point infinetely many times. Simeon === Subject: Re: Looking for a surjection or R^2 Simeon Stefanov a .8ecrit : > What about f(x,y) = e^(iy)(e^x+1) + y? It's easy to check that the > Jacobian is > 0 Not so easy :-) But right, it is e^x(e^x+1-sin y) and f is not bijective: f(a,pi) = f(b,3pi) = 0 for some > a, b. The sujectivity seems OK from some geometrical evidence: the map > e^(iy)(e^x+1) winds the plane onto the complement of the unit circle, > and adding y to it, will cover each point infinetely many times. But this is not so clear...Ok, I will check. But I still prefer the answer of The World Wide Wade. > Simeon === Subject: Re: Looking for a surjection or R^2 <448a96a0$0$7770$7a628cd7@news.club-internet.fr Simeon Stefanov a .8ecrit : > What about f(x,y) = e^(iy)(e^x+1) + y? It's easy to check that the > Jacobian is > 0 > Not so easy :-) But right, it is e^x(e^x+1-sin y) > and f is not bijective: f(a,pi) = f(b,3pi) = 0 for some > a, b. The sujectivity seems OK from some geometrical evidence: the map > e^(iy)(e^x+1) winds the plane onto the complement of the unit circle, > and adding y to it, will cover each point infinetely many times. > But this is not so clear... O.K., I'll try to show the surjectivity of f(x,y). We have to prove that for any (u,v) in R^2 the system u = cos(y)(e^x+1) + y v = sin(y)(e^x+1) has a solution. Let first v<>0, that is equivalent to sin(y)<>0. Then we get from the system (*) u = ctan(y)v + y. Suppose now that y belongs to (0, pi) and x is in R. Then v is in (0,+infty) and since ctan(y) is taking any value from R decreasing monotonely, as y in (0, pi), we get from (*) that any point (u,v) with v > 0 is in the image of the map f. Similarly, if y is in (pi, 2pi), v is in (0,-infty) and we get as above that any point (u,v) with v<0 is covered by f. Let finally v = 0 and u is arbitrary. Now, if u>=0, set y = -2pi. Then the system gives u = e^x + 1 - 2pi, which has a solution for some x. If u<0, set y=pi. Then u = -e^x - 1 + pi, that has a solution again. Hence the image of f(x,y) is R^2. Simeon === Subject: Re: Looking for a surjection or R^2 , > Denis Feldmann with jacobian everywhere not zero. My best example so far is an > holomorphic function (with the obvious isomorphism between C and R^2) > like the antiderivative of exp(z^2) (which is surjective by Picard > theorem), but I would like something simpler, more explicit, and where > the proof of surjectivity uses only elementary calculations... > Let f be any smooth function on R such that f(y) = 0 if y <= 0, > f' <= 0 on [0,Pi], f' >= 0 on [Pi,2Pi], f(2Pi) < 0, and f(y) = > f(2Pi) for y >= 2Pi. Writing z = x + iy, define F(z) = e^z + > f(y). Then the Jacobian of F at x + iy is e^(2x) - > e^x*sin(y)*f'(y), which = 0 iff e^x = sin(y)*f'(y). The right > hand side is always <= 0, so the equation cannot be satisfied. > Therefore the Jacobian of F is never 0. F is surjective because > below the x-axis it maps onto C {0}, Here I should have said that below the x_axis, F(z) = e^z, where it maps many to one onto C {0}. > and on the line y = 2Pi it > is the function e^x + f(2Pi); because f(2Pi) < 0, this function > takes on the value 0. And finally, F is not a bijection because of its behavior below the x-axis. === Subject: Re: Looking for a surjection or R^2 , > Denis Feldmann I am looking for a smooth surjection of R^2 onto itself, not bijective, > with jacobian everywhere not zero. My best example so far is an > holomorphic function (with the obvious isomorphism between C and R^2) > like the antiderivative of exp(z^2) (which is surjective by Picard > theorem), but I would like something simpler, more explicit, and where > the proof of surjectivity uses only elementary calculations... > Let f be any smooth function on R such that f(y) = 0 if y <= 0, > f' <= 0 on [0,Pi], f' >= 0 on [Pi,2Pi], f(2Pi) < 0, and f(y) = > f(2Pi) for y >= 2Pi. Writing z = x + iy, define F(z) = e^z + > f(y). Then the Jacobian of F at x + iy is e^(2x) - > e^x*sin(y)*f'(y), which = 0 iff e^x = sin(y)*f'(y). The right > hand side is always <= 0, so the equation cannot be satisfied. > Therefore the Jacobian of F is never 0. F is surjective because > below the x-axis it maps onto C {0}, > Here I should have said that below the x_axis, F(z) = e^z, where > it maps many to one onto C {0}. > and on the line y = 2Pi it > is the function e^x + f(2Pi); because f(2Pi) < 0, this function > takes on the value 0. > And finally, F is not a bijection because of its behavior below > the x-axis. who he is? === Subject: Re: Looking for a surjection or R^2 > , > Denis Feldmann I am looking for a smooth surjection of R^2 onto itself, not bijective, > with jacobian everywhere not zero. My best example so far is an > holomorphic function (with the obvious isomorphism between C and R^2) > like the antiderivative of exp(z^2) (which is surjective by Picard > theorem), but I would like something simpler, more explicit, and where > the proof of surjectivity uses only elementary calculations... > > Let f be any smooth function on R such that f(y) = 0 if y <= 0, > f' <= 0 on [0,Pi], f' >= 0 on [Pi,2Pi], f(2Pi) < 0, and f(y) = > f(2Pi) for y >= 2Pi. Writing z = x + iy, define F(z) = e^z + > f(y). Then the Jacobian of F at x + iy is e^(2x) - > e^x*sin(y)*f'(y), which = 0 iff e^x = sin(y)*f'(y). The right > hand side is always <= 0, so the equation cannot be satisfied. > Therefore the Jacobian of F is never 0. F is surjective because > below the x-axis it maps onto C {0}, > Here I should have said that below the x_axis, F(z) = e^z, where > it maps many to one onto C {0}. > and on the line y = 2Pi it > is the function e^x + f(2Pi); because f(2Pi) < 0, this function > takes on the value 0. > And finally, F is not a bijection because of its behavior below > the x-axis. Yes. === Subject: Re: Looking for a surjection or R^2 The World Wide Wade a .8ecrit : > , >> Denis Feldmann > I am looking for a smooth surjection of R^2 onto itself, not bijective, > with jacobian everywhere not zero. My best example so far is an > holomorphic function (with the obvious isomorphism between C and R^2) > like the antiderivative of exp(z^2) (which is surjective by Picard > theorem), but I would like something simpler, more explicit, and where > the proof of surjectivity uses only elementary calculations... >> Let f be any smooth function on R such that f(y) = 0 if y <= 0, >> f' <= 0 on [0,Pi], f' >= 0 on [Pi,2Pi], f(2Pi) < 0, and f(y) = >> f(2Pi) for y >= 2Pi. Writing z = x + iy, define F(z) = e^z + >> f(y). Then the Jacobian of F at x + iy is e^(2x) - >> e^x*sin(y)*f'(y), which = 0 iff e^x = sin(y)*f'(y). The right >> hand side is always <= 0, so the equation cannot be satisfied. >> Therefore the Jacobian of F is never 0. F is surjective because >> below the x-axis it maps onto C {0}, > Here I should have said that below the x_axis, F(z) = e^z, where > it maps many to one onto C {0}. >> and on the line y = 2Pi it >> is the function e^x + f(2Pi); because f(2Pi) < 0, this function >> takes on the value 0. > And finally, F is not a bijection because of its behavior below > the x-axis. > Yes. guide us towards a completely explicit (C-infinite) function, namely F(x,y)= (sin(y)(e^x - 1/2), -cos(y) e^x) . I will try to give proper credits to it if we happen to publish it (in a course for French teachers) === Subject: Re: Looking for a surjection or R^2 >> I am looking for a smooth surjection of R^2 onto itself, not bijective, >> with jacobian everywhere not zero. ... >guide us towards a completely explicit (C-infinite) function Here's the example with irregular branched coverings that I promised. *I* think it's already completely explicit as I'm about to give it; but further explicitation can be done (and will be sketched) at the expense of simplicity. As before, identify R^2 with C, and let p(z)=z^3-3z. Then p is a 3-sheeted irregular branched cover of C over C, with critical points 1 and -1, and critical values -2 = p(1) = p(-2) and 2 = p(-1) = p(2). The preimage p^{-1}(R), call it G, is the union of R and two (real) parabolas, forming a figure approximately like this: _____/__ / in which the 2 doublepoints of G are the critical points of p. Note that C G the union of 6 open 2-cells, on each of which the restriction of p is a diffeomorphism onto either the upper or the lower half-plane; in fact the restriction of p to the closure of the open 2-cell is a homeomorphism onto the closed half-plane, which straightens out the right angle(s) on the boundary but is otherwise smooth. Let K be the closure of the bottom 2-cell, that is, the one that contains the negative imaginary axis, and U its complement. Then p has no critical points on U, p(U) = C, and p|U is not injective. Since U is manifestly diffeomorphic to C, we're done, by my standards. To bring the example up to what *may* be your more exacting standards, we need only write down a formula for a diffeomorphism of U to C. There's lots of ways Lee Rudolph === Subject: Re: Looking for a surjection or R^2 ... >As before, identify R^2 with C, and let p(z)=z^3-3z. Then p is a >3-sheeted irregular branched cover of C over C, with critical points >1 and -1, and critical values -2 = p(1) = p(-2) and 2 = p(-1) = p(2). >The preimage p^{-1}(R), call it G, is the union of R and two (real) >parabolas which is of course wrong--with z = x+iy as usual, G is the zero-set of y(3x^2-y^2-1), so it's the union of R and the two branches of a *hyperbola*. Lee Rudolph === Subject: Re: Looking for a surjection or R^2 Lee Rudolph a .8ecrit : > ... >> As before, identify R^2 with C, and let p(z)=z^3-3z. Then p is a >> 3-sheeted irregular branched cover of C over C, with critical points >> 1 and -1, and critical values -2 = p(1) = p(-2) and 2 = p(-1) = p(2). >> The preimage p^{-1}(R), call it G, is the union of R and two (real) >> parabolas > which is of course wrong--with z = x+iy as usual, G is the zero-set > of y(3x^2-y^2-1), so it's the union of R and the two branches of a > *hyperbola*. Are you trying to coopt JSH in this thread? > Lee Rudolph === Subject: Polytopes I'm reading a paper and it says We define polyhedral gauges according to Minkowski (1911): Defnition 2.1 Let B be a polytope in R^n containing the origin in its interior and let z in R^n. (1) The polyhedral gauge g:R^n -> R of z is defined as g(z) := min{L >= 0 | z in L*B} (2) If B is symmetric with respect to the origin, then g is called a block norm. (3) The vectors defined by the extreme points of the unit ball B of g are called fundamental vectors and are denoted by v^i. The fundamental vectors defined by the extreme points of a facet of B span a fundamental cone. (4) A block norm g with a unit ball B is called oblique (Schandl et al., 2001a) if it has the following properties: (i) g is absolute, i.e., g(w) = g(u) for all w in R(u) := {w in R^n | |w_i| = |u_i| for all i = 1,..., n}, (ii) (z ? R^n_{>=}) intersection R^n_{>=} intersection Boundary of B = {z} for all z in (Boundary of B intersect R^n_{>=}) (I copyed and pasted and a lot of stuff didn't come out right. I think I fixed it though.) I'm not sure I even understand what a polytope is. I've looked on the net but there are several definitions that are somewhat different. It seems that g is defined as the minimumly sized polytope that will contain z? Why is this useful? in (3) it says that The vectors defined by the extreme points of the unit ball B of g are called fundamental vectors and are denoted by v^i the extreme points? Its basicaly talking abuot the boundary of B? (since all points on the surface of B are extreme) But then what the heck does v^i mean? Any ideas? Jon === Subject: Re: Polytopes BTW, whats R^n_{>=}? (i.e., R^n with a >= sign as a subscript)? === Subject: A question in book Concrete Mathematics In page 15, ... Suppose we now relax the radix 2 notation to allow arbitrary digits instead of just 0 and 1. The derivation above tell us that f(...) = (...) (1.16) ... I don't understant how the equation 1.16 come out, what does relax the radix 2 notation mean? Please help me. === Subject: Re: A question in book Concrete Mathematics > In page 15, > ... > Suppose we now relax the radix 2 notation to allow arbitrary digits > instead > of just 0 and 1. The derivation above tell us that > f(...) = (...) (1.16) > ... > I don't understant how the equation 1.16 come out, > what does relax the radix 2 notation mean? > Please help me. Instead of using just digits 0 and 1, allow any digits. So 1.16 would mean 1 + 1/2 + 6/4 [Or is 1.16 the label on the equation? Does 1.15 appear on the previous labeled display?] === Subject: Re: A question in book Concrete Mathematics A N Niel a .8ecrit : >> In page 15, >> ... >> Suppose we now relax the radix 2 notation to allow arbitrary digits >> instead >> of just 0 and 1. The derivation above tell us that >> f(...) = (...) (1.16) >> ... >> I don't understant how the equation 1.16 come out, >> what does relax the radix 2 notation mean? >> Please help me. > Instead of using just digits 0 and 1, allow any digits. So > 1.16 would mean 1 + 1/2 + 6/4 > [Or is 1.16 the label on the equation? Does 1.15 appear on the > previous labeled display?] In fact, (1.6) is indeed the number of hte equation, but it turns out your explanation is exactly what is intended by the autors, i.e. to interpret [301.16]_2 as 3*4+0*2+1*1+1*1/2+6*1/4. By the way, the pertinent marginal graffiti here is 'relax' = 'destroy'. === Subject: Re: A question in book Concrete Mathematics > Suppose we now relax the radix 2 notation to allow arbitrary digits > instead of just 0 and 1. The derivation above tell us that > f(...) = (...) (1.16) > ... > I don't understant how the equation 1.16 come out, > what does relax the radix 2 notation mean? I don't understand how you can expect us to answer. Suppose we know nothing about your book and its context. === Subject: Re: A question in book Concrete Mathematics >> Suppose we now relax the radix 2 notation to allow arbitrary digits >> instead of just 0 and 1. The derivation above tell us that >> f(...) = (...) (1.16) >> ... >> I don't understant how the equation 1.16 come out, >> what does relax the radix 2 notation mean? > I don't understand how you can expect us to answer. > Suppose we know nothing about your book and its context. Very simple.. then don't reply. He's not asking if you know or not... he's asking if you do know then could you please help. You need to stop trying to be a jerk. You don't have to reply to every post with some smart ass comment. If you can answer the question(in atleast a decent manner) then do so... if you can't then don't reply. Very simple and even then genius that you are can figure that out. === Subject: Re: A question in book Concrete Mathematics <128lgbchjl7h071@corp.supernews.com I don't understand how you can expect us to answer. > Suppose we know nothing about your book and its context. > Very simple.. then don't reply. He's not asking if you know or not... he's > asking if you do know then could you please help. You need to stop trying > to be a jerk. You don't have to reply to every post with some smart ass > comment. If you can answer the question(in atleast a decent manner) then do > so... if you can't then don't reply. Very simple and even then genius that > you are can figure that out. Gee whiz Mr. A.D. I may call you A.D may I not, for our familiarity? Anyway, Mr. A.D., I fail in this instance to see how your Dissonance is Abstract. A minor laspe no doubt soon to be quickly and easily corrected. === Subject: Re: Nicer proof in basic topology? <4489A099.6030800@netscape.net>Early on (I.13), Kelley states the theorem that an uncountable subset >>of a second countable space must contain an accumulation point. >>Let G be a countable base and let A be an uncountable subset of >>the space. Let C = U{B in G: int(B,A) is a singleton}. >>Since int(C,A) is countable, AC is nonempty. >Why is C / A countable? > Let H = {B in G: int(B,A) is a singleton. For each B in H, there > is a unique element a_B in int(B,A). C = UH = {a_B: B in H}. H is > countable. (I was hoping the reader could do that him/herself.) You claim an injection f:H -> C / A C / A = /{ B / A | B in G, B / A singleton } |C / A| <= sum_H 1 <= sum_G 1 <= aleph_0 >>For each a in AC and each B in G >>such that a in B, B intersects A{a}. Q.E.D. Nice. Do you not have stronger result? Namely uncountably many elements of A are limit points of A. >You or Kelley is confusing accumulation point with limit point >which are equivalent in T1 spaces. > Kelley defines an accumulation point of a set A as a point x such > that each open set which contains x intersects A {x}. He is not alone. Nor topology the only field that requires translations between authors. Are all spaces to Kelley Hausdorff? === Subject: An Optimization problem hello, I had some problem regarding the formulation of a function regarding the following problem. I had two variables, say Et (representing the total energy consumed during a packet transfer from one point in the network to another point) and Er (Remaining or current available energy of a node). Now i want to relate these two quantities in such a way that value of the function (that describes the relation between these two quantities) gives me the best possible combination. Remember that i would like to adopt set of values in which Er have maximum value and Et has minimum value. Value of Er may have an initial value of 'm' after which it will continue to decrease till zero. Et has one constraint and that is: Et > 0; saleem === Subject: Re: An Optimization problem > hello, > I had some problem regarding the formulation of a function regarding > the following problem. > I had two variables, say Et (representing the total energy consumed > during a packet transfer from one point in the network to another > point) and Er (Remaining or current available energy of a node). > Now i want to relate these two quantities in such a way that value of > the function (that describes the relation between these two quantities) > gives me the best possible combination. Remember that i would like to > adopt set of values in which Er have maximum value and Et has minimum > value. > Value of Er may have an initial value of 'm' after which it will > continue to decrease till zero. Et has one constraint and that is: Et 0; > saleem What your looking for is called Multi Attribute Utility Theory. You do not have to create a function that relates your two quantities in one. If you do that then you are basically still in the real of single attribute utility theory. You have to come up with the function that does that yourself though. I'm still learning about the MAUT myself so I can't help to much. There are plety of algorithms out there but I have yet to find any resonable introduction to the theory and everythign I have found is pretty much research stuff. You could also look for geometric programming and other theories like that(there are a bunch). Unfortunately I doubt you will find a quick answer but if you do let me know too. I'v been looking and reading for the last 1-2 weeks on this type of stuff trying to figure out how I can use it to solve my problem. Most everything I have read talks about the pareto set and front and such... lots of strange concepts that I have never heard of before ;/ Note that you can simply use single attribute utility theory to do what you want. This was my natural response to solving my problem which is similar(except I have many more functions to optimize). the idea is imply to create a weight sum of some functional forms of your quantities. i.e. sum(w_i*f_i(Q_i),i=1..n) Where w_i is the weight of the ith quanity, f_i is a utility function that fixes up the ith quantity Q_i. I gave up on this approach because I have many quantities and I couldn't figure out any easy way to deal with findind w_i and f_i. I was just reading a paper that talked about a special case when dealing with only two quantities. IT had something to do with the nadir pioint. Not sure if that helps but thats the best I can do. Jon === Subject: Re: An Optimization problem <128lh1lt581vt3b@corp.supernews.com> u. saleem > hello, > I had some problem regarding the formulation of a function regarding > the following problem. > I had two variables, say Et (representing the total energy consumed > during a packet transfer from one point in the network to another > point) and Er (Remaining or current available energy of a node). > Now i want to relate these two quantities in such a way that value of > the function (that describes the relation between these two quantities) > gives me the best possible combination. Remember that i would like to > adopt set of values in which Er have maximum value and Et has minimum > value. > Value of Er may have an initial value of 'm' after which it will > continue to decrease till zero. Et has one constraint and that is: Et 0; > saleem > What your looking for is called Multi Attribute Utility Theory. > You do not have to create a function that relates your two quantities in > one. If you do that then you are basically still in the real of single > attribute utility theory. You have to come up with the function that does > that yourself though. > I'm still learning about the MAUT myself so I can't help to much. There are > plety of algorithms out there but I have yet to find any resonable > introduction to the theory and everythign I have found is pretty much > research stuff. > You could also look for geometric programming and other theories like > that(there are a bunch). > Unfortunately I doubt you will find a quick answer but if you do let me know > too. I'v been looking and reading for the last 1-2 weeks on this type of > stuff trying to figure out how I can use it to solve my problem. > Most everything I have read talks about the pareto set and front and such... > lots of strange concepts that I have never heard of before ;/ > Note that you can simply use single attribute utility theory to do what you > want. This was my natural response to solving my problem which is > similar(except I have many more functions to optimize). > the idea is imply to create a weight sum of some functional forms of your > quantities. > i.e. > sum(w_i*f_i(Q_i),i=1..n) > Where w_i is the weight of the ith quanity, f_i is a utility function > that fixes up the ith quantity Q_i. > I gave up on this approach because I have many quantities and I couldn't > figure out any easy way to deal with findind w_i and f_i. > I was just reading a paper that talked about a special case when dealing > with only two quantities. IT had something to do with the nadir pioint. > Not sure if that helps but thats the best I can do. > Jon === Subject: Re: Internal Bisectors of a Triangle > Maury Barbato nous a r.8ecemment amicalement > signifi.8e : > I have > now two more elementary questions: (I) given > three distinct straight > lines concurrent in a > point I, is there a triangle having these lines > as > internal bisectors? > > Yes, as soon as the angles made by the three > lines are all greater than > pi/2 > > In fact, you just have to notice the following > point : > Given a triangle ABC and the intersection I of > angles bissectors : > angle AIB = pi/2 + angle ACI > angle BIC = pi/2 + angle BAI > angle CIA = pi/2 + angle CBI > Yes, this affords a rather simpler construction > than the one I > suggested at alt.math.undergrad! > > (II) if such a triangle exists, is it > essentially unique > (that is every other triangle which solves the > problem > can be obtained by the first by a homothety > with center > O)? > > Yes, if you accept also homothety with n.8egative > -- > Patrick > Suggested marking off given proportions of AI, BI, CI > ( which can be > calculated or obtained by Ruler/Compass) in : > se_frm/thread/9634317e4e965288/0ff06494eba11b9b?lnk=ra > ot#0ff06494eba11b9b > Shall appreciate comments. Did not expect ex-circles > could be also > accomodated in the solution.Is it what you meant by > negative k > homothety ? > Narasimham Ex-circles? How did it cross your mind? We didn't mention in-circles nor ex-circles. Homothety with center O and ratio k: if O is a point in the euclidean plane and k is a real number, then the image of a point P is the point Q such that the following vectorial identity holds: OQ=kOP. Maury === Subject: Re: Internal Bisectors of a Triangle schrieb Maury Barbato : >> Maury Barbato: >> I have now two more elementary questions: >> (I) given three distinct straight >> lines concurrent in a point I, >> is there a triangle having these lines as >> internal bisectors? >> Yes, as soon as the angles made by the three >> lines are all greater than pi/2 >> In fact, you just have to notice the following point : >> Given a triangle ABC >> and the intersection I of angles bissectors : >> angle AIB = pi/2 + angle ACI >> angle BIC = pi/2 + angle BAI >> angle CIA = pi/2 + angle CBI >> Yes, this affords a rather simpler construction than the one I >> suggested at alt.math.undergrad! >> (II) if such a triangle exists, is it essentially unique >> (that is every other triangle which solves the problem >> can be obtained by the first by a homothety with center I)? >> Yes, if you accept also homothety with negative k [ ... ] >> Shall appreciate comments. Did not expect ex-circles could be also >> accomodated in the solution. [ ... ] > Ex-circles? How did it cross your mind? We didn't mention > in-circles nor ex-circles. [ ... ] Well,you mentioned that the point I shall become the _incenter_ of the triangle ABC. In fact, the following construction yields a triangle ABC for which the given lines x,y,z that are concurrent in I, are either all interior angle bisectors, and the point I is the incenter of ABC, or the lines are bisectors, but possibly of outer angles of the triangle ABC, and the point I is an excenter of ABC: Pick a point A =/= I on x. Reflect A on the lines y, resp. z - call the image points A', resp. A''. The interesection point of the line A'A'' with y is (called) B; the intersection with z is (called) C. The triangle ABC is unique up to homotheties when none of the lines x, y, z are perpendicular to each other. If the point I is contained inside the triangle ABC, it will be its incenter; if not, it will be one of the excenters of ABC. === Subject: Re: Internal Bisectors of a Triangle > However, I didn't yet understand what are the > answer to > my questions. > > (I) Given three arbitrary lenghts, is there a > triangle > whose angle bisectors have the given lengths? > > (II) Is it unique? > > My Best Regsrds, > Maury > > Just to clarify, by the length of an angle > bisector > do you mean the > distance from the vertex to the intersection of > the > bisector with the > opposite side of the triangle? (Rather than, for > example, the distance > from the vertex to the point of intersection of > the > bisectors.) > Surely the length of an angle bisector is the > distance > from the vertex to the intersection of the bisector > with > the opposite side of the triangle. > Right. > I have studied Chisini's work. I'm quite sure now > that > the answer to (II) is no. > I find that rather surprising. Can you provide an > example of two > different triangles with the same length angle > bisectors? (Excluding > mirror images, which I would count as the same > triangle.) >Maybe, (I) has a positive > answer, but I'm not sure. > My feeling was that the answer to (I) is yes (but > my feeling was that > the answer to (II) is also yes, so if that's wrong > then my feelings > are obviously not worth so much!) As I have already written, the algebraic translation of the geometric problem made by Chisini is an algebraic equation in two variables of degree 10. This fact doesn't ensure the uniqueness of the solution nor its existence. So, if you believe that my questions have positive answers, I'm happy. But without proofs ... Maury === Subject: Re: Internal Bisectors of a Triangle <23464520.1149936692075.JavaMail.jakarta@nitrogen.mathforum.org > However, I didn't yet understand what are the > answer to > my questions. > > (I) Given three arbitrary lenghts, is there a > triangle > whose angle bisectors have the given lengths? > > (II) Is it unique? > > My Best Regsrds, > Maury > > Just to clarify, by the length of an angle > bisector > do you mean the > distance from the vertex to the intersection of > the > bisector with the > opposite side of the triangle? (Rather than, for > example, the distance > from the vertex to the point of intersection of > the > bisectors.) > > Surely the length of an angle bisector is the > distance > from the vertex to the intersection of the bisector > with > the opposite side of the triangle. > Right. > I have studied Chisini's work. I'm quite sure now > that > the answer to (II) is no. > I find that rather surprising. Can you provide an > example of two > different triangles with the same length angle > bisectors? (Excluding > mirror images, which I would count as the same > triangle.) >Maybe, (I) has a positive > answer, but I'm not sure. > My feeling was that the answer to (I) is yes (but > my feeling was that > the answer to (II) is also yes, so if that's wrong > then my feelings > are obviously not worth so much!) > As I have already written, the algebraic translation > of the geometric problem made by Chisini is > an algebraic equation in two variables of degree 10. > This fact doesn't ensure the uniqueness of the solution > nor its existence. So, if you believe that my questions > have positive answers, I'm happy. But without proofs ... > Maury You said you were quite sure that the answer to (II) is no, which I understood to mean certain. But in fact you're saying you're *not* certain that the answer is no? === Subject: Re: Internal Bisectors of a Triangle > > However, I didn't yet understand what are > the > answer to > my questions. > > (I) Given three arbitrary lenghts, is there > triangle > whose angle bisectors have the given > lengths? > > (II) Is it unique? > > My Best Regsrds, > Maury > > Just to clarify, by the length of an angle > bisector > do you mean the > distance from the vertex to the intersection > of > the > bisector with the > opposite side of the triangle? (Rather than, > for > example, the distance > from the vertex to the point of intersection > of > the > bisectors.) > > Surely the length of an angle bisector is the > distance > from the vertex to the intersection of the > bisector > with > the opposite side of the triangle. > > Right. > > I have studied Chisini's work. I'm quite sure > now > that > the answer to (II) is no. > > I find that rather surprising. Can you provide an > example of two > different triangles with the same length angle > bisectors? (Excluding > mirror images, which I would count as the same > triangle.) > >Maybe, (I) has a positive > answer, but I'm not sure. > > My feeling was that the answer to (I) is yes > (but > my feeling was that > the answer to (II) is also yes, so if that's > wrong > then my feelings > are obviously not worth so much!) > As I have already written, the algebraic > translation > of the geometric problem made by Chisini is > an algebraic equation in two variables of degree > 10. > This fact doesn't ensure the uniqueness of the > solution > nor its existence. So, if you believe that my > questions > have positive answers, I'm happy. But without > proofs ... > Maury > You said you were quite sure that the answer to > (II) is no, which I > understood to mean certain. But in fact you're > saying you're *not* > certain that the answer is no? No, I'm not sure. The only fact that induced me to suspect that the answer is no is that the mentioned equation has degree 10, nothing else. Maury === Subject: Re: Does the Calculus rest on Euclid? On Fri, 09 Jun 2006 11:27:19 -0400, Hatto von Aquitanien >> On Thu, 08 Jun 2006 09:26:29 -0400, Hatto von Aquitanien >> On Wed, 07 Jun 2006 13:44:23 -0400, Hatto von Aquitanien >I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I >have never felt satisfactorily convinced that the transition from the >Riemann sum approximation to a smooth curve is logically founded upon >axioms I have assumed at the outset. These axioms are those of formal >logic and those of Euclid. >> Well, those are not the axioms that are used in the >> standard approach to a rigorous treatment of calculus. >I've checked several so-called advanced treatments of analysis, and to be >quite honest, they don't seem to state their assumptions very clearly. I >will concede that there is no need to appeal to Euclid in the calculus of >single variable functions. Nonetheless, I do believe there is an >assumption of continuity which is geometric in nature, and it is at the >foundation of all our mathematical reasoning. >> Believe what you want. It's not so. You start with the integers, >> construct the rationals from them. Now you define the reals as >> follows: >> A real number is a set S of rationals with the following properties: >> (i) S is non-empty >> (ii) S does not contain every rational >> (iii) If s, t are rational, s is in S and t > s, then t is in S >> (iv) S has no smallest element. >Pi is not a member of the set of rationals. No, it's not. The definition above does not require that every real number be rational. A real number _is_ a certain _set_ of rationals. > I do not accept your definition >of the real numbers. Guffaw. It's not my definition, it's a version of one of the standard definitions. Saying you do not accept a _definition_ is silly. Yes, by the definition above there _is_ a real number commonly called pi. And it's irrational. >> No assumptions of continuity in sight. >To say that you can order things implies a concept of a continuum. What? Given the definition above, we _define_ x < y for real numbers x, y by saying x < y if and only if y is a proper subset of x. >implication, those of axiomatic set theory which is isomorphic to symbolic >logic. >> Uh no, set theory is symbolic logic _plus_ some axioms. >Which can be stated in terms of boolean algebra. Uh, no. Don't you get the feeling that just _maybe_ you should _learn_ something about these things, before proclaiming that mathemtaticians are all wrong? > One might haggle over >whether those boolean algebra statements implicitly assume ideas from set >theory. I will concede that point. > In differential geometry it is not infrequently stated that every >n-dimensional non-Euclidian space is described in terms of an n+m >Euclidian embedding space. >> Uh, no. It's stated that every compact n-dimensional manifold can be >> embedded in R^(n+m) for some m. And then it's _proved_ that this is >> the case. >Levi-Civita's development is interesting. He sure likes that term >'a-priori'. > Where m is a positive integer such that m >= 1. Now, >that usually appears in the context of discussing physics, so there may be >a certain amount of poetic license taking place. >> Um. Possibly you've _seen_ this in the context of physics. >> If your knowledge of mathematics comes from expositions >> in the context of physics it's not surprising that you >> think that things are not justified as well as they should >> be - physicists tend to just take mathematicians' word >> for a lot of the finicky details. Find a _math_ book >> on differential geometry and you'll find a _proof_ of >> the theorem above (or not, in which case you need to >> find a different book.) >What is the geometry of a locally tangent space? > For example, I do not >know if it would be more correct to say the embedding space is Lorentzian >in the case of general relativity. >Nonetheless, I believe when all is said and done, we cannot reason about >any >of these ideas without appeal to our innate Euclidian space. The >developments I have seen for the real numbers beginning with Peano's >axioms seem to take a step of faith either explicitly through the >continuum hypothesis, or implicitly when they jump off the firm ground of >the rational numbers using the demonstrably irrational numbers as a >justification for the existence of a continuum of real numbers. >> Have you ever seen a careful exposition of the basic properties >> of reals, defined in terms of Dedekind cuts? >Every discussion I have seen of this topic requires a concept of completed >infinity, and a notion of infinite repeatability. I have no concept of >repeatability without a concept of time. My concept of time is a-priori >continuous. Likewise for my concept of space. Describe the fundamental >axioms upon which these developments are based without appeal to words >dealing with space or time. ************************ David C. Ullrich === Subject: Re: Does the Calculus rest on Euclid? > On Fri, 09 Jun 2006 11:27:19 -0400, Hatto von Aquitanien > A real number is a set S of rationals with the following properties: >> (i) S is non-empty > (ii) S does not contain every rational > (iii) If s, t are rational, s is in S and t > s, then t is in S > (iv) S has no smallest element. >>Pi is not a member of the set of rationals. > No, it's not. The definition above does not require that > every real number be rational. A real number _is_ a certain > _set_ of rationals. >> I do not accept your definition >>of the real numbers. > Guffaw. It's not my definition, it's a version of one of > the standard definitions. Saying you do not accept > a _definition_ is silly. > Yes, by the definition above there _is_ a real number > commonly called pi. And it's irrational. If a real number is a set of rationals, and there is only one real number designated by the concept indicated by the phrase a real number, then, in the case of pi either the set called pi is identical to the object called pi, or we are avoiding (or more correctly, evading) the original topic. > No assumptions of continuity in sight. >>To say that you can order things implies a concept of a continuum. > What? > Given the definition above, we _define_ x < y for real numbers > x, y by saying x < y if and only if y is a proper subset of x. But you cannot communicate any of that with out some appeal to my a priori sense of continuum. The very simple fact that I see these symbols ordered on my computer screen involves my a priori concept of continuum. >implication, those of axiomatic set theory which is isomorphic to >>symbolic logic. >> Uh no, set theory is symbolic logic _plus_ some axioms. >>Which can be stated in terms of boolean algebra. > Uh, no. > Don't you get the feeling that just _maybe_ you should _learn_ > something about these things, before proclaiming that mathemtaticians > are all wrong? Perhaps 'boolean algebra' may not be the best term to use. The distinctions between 'boolean algebra', formal logic, propositional logic, etc. are not always clearly maintained in computer science literature, and I would need to carefully review the literature to draw clear distinctions. What I am saying is that set theory begins with concepts which are identical to those found expressed in terms of the primary logical connectives. http://baldur.globalsymmetry.com/open-source/org/sth/math/logic.xhtml It's really quite simple. If your system is logically consistent and sufficiently powerful, then it provides a means of transitioning from any given state (collection of statements forming an aggregate statement) to another state defined by the rules for manipulating symbols. Any such system can be codified in binary. >>What is the geometry of a locally tangent space? -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? <8ddl829t0pji2ecj4apokibb9gvgjdpflm@4ax.com> designated by the concept indicated by the phrase a real number, then, in > the case of pi either the set called pi is identical to the object called > pi, or we are avoiding (or more correctly, evading) the original topic. It doesn't make sense to talk about what the object called pi is. Numbers don't exist in the same sense as the pen sitting on my desk. Certain properties characterize pi, and especially its properties as a member of the set of all real numbers. In different models of the real numbers, pi might be a set of rational numbers, an equivalence class of sequences of rational numbers, or even a point on a line. Whether or not these objects are identical is irrelevant. The point is that all these models behave equivalently in the ways that are significant for real numbers. What color is pi? It doesn't matter. === Subject: Re: Does the Calculus rest on Euclid? >> If a real number is a set of rationals, and there is only one real >> number designated by the concept indicated by the phrase a real number, >> then, in the case of pi either the set called pi is identical to the >> object called pi, or we are avoiding (or more correctly, evading) the >> original topic. > It doesn't make sense to talk about what the object called pi is. > Numbers don't exist in the same sense as the pen sitting on my desk. > Certain properties characterize pi, and especially its properties as a > member of the set of all real numbers. In different models of the real > numbers, pi might be a set of rational numbers, an equivalence class of > sequences of rational numbers, or even a point on a line. Whether or > not these objects are identical is irrelevant. The point is that all > these models behave equivalently in the ways that are significant for > real numbers. What color is pi? It doesn't matter. Well, at the risk of posting on low blood caffeine, I believe it does matter. Again, this is a topic to which I simply cannot direct my full attention at present. What appears to be happening is a crossing of levels of abstraction. This is how Russell tried to climb out of his paradox. My instincts tell me that there is a contradiction to be found by reviewing carefully all the statements and assumptions. Or perhaps there is a necessary step missing which, if completed, would result in a contradiction. I am confident that at least one concept has been accepted from the properties I ascribe to the continuum in the development of the real numbers. That is the concept of completed infinity. OTOH, since iterations are discrete, one can argue that they are countably infinite. This really wasn't what I had in mind when I started this thread. I was really trying to get a handle on a rather elementary idea in geometry. That concept is the idea of measuring the length of the arc subtending a radial angle in a unit circle. I may have been well advised not to have mentioned the calculus. I could see no way of going from the notion of an n-sided polygonal approximation to the circle (or arc thereof) without invoking the Powers on High. I was thinking in terms of Euclid, but my reasoning was following the line typically used in arriving at the arclength of a curve. Since virtually all developments I have seen for the arclength theorem invoke geometric analogs, its hard for me to identify the which is purely analytical. -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? On Mon, 12 Jun 2006 13:17:39 -0400, Hatto von Aquitanien >[...] >Well, at the risk of posting on low blood caffeine, I believe it does >matter. Again, this is a topic to which I simply cannot direct my full >attention at present. What appears to be happening is a crossing of levels >of abstraction. This is how Russell tried to climb out of his paradox. My >instincts tell me that there is a contradiction to be found by reviewing >carefully all the statements and assumptions. Or perhaps there is a >necessary step missing which, if completed, would result in a >contradiction. You'd really sound a lot less, um, I'll let you supply the adjective - you'd sound a lot less something if you found the contradiction first and then told us about it. See, we all know what contradictions are, and we're against them. Some of us _do_ have the time to devote to these things, and we've devoted a lot of time to them for many years. Without finding any contradictions. >I am confident that at least one concept has been accepted from the >properties I ascribe to the continuum in the development of the real >numbers. That is the concept of completed infinity. OTOH, since >iterations are discrete, one can argue that they are countably infinite. >This really wasn't what I had in mind when I started this thread. I was >really trying to get a handle on a rather elementary idea in geometry. >That concept is the idea of measuring the length of the arc subtending a >radial angle in a unit circle. Whoever told you that that was elementary lied. > I may have been well advised not to have >mentioned the calculus. >I could see no way of going from the notion of an n-sided polygonal >approximation to the circle (or arc thereof) without invoking the Powers >on High. I was thinking in terms of Euclid, but my reasoning was following >the line typically used in arriving at the arclength of a curve. Since >virtually all developments I have seen for the arclength theorem invoke >geometric analogs, its hard for me to identify the which is purely >analytical. ************************ David C. Ullrich === Subject: Re: Does the Calculus rest on Euclid? <8ddl829t0pji2ecj4apokibb9gvgjdpflm@4ax.com> <18OdnfOK3akkPBDZnZ2dnUVZ_v6dnZ2d@speakeasy.net> If a real number is a set of rationals, and there is only one real >> number designated by the concept indicated by the phrase a real number, >> then, in the case of pi either the set called pi is identical to the >> object called pi, or we are avoiding (or more correctly, evading) the >> original topic. > It doesn't make sense to talk about what the object called pi is. > Numbers don't exist in the same sense as the pen sitting on my desk. > Certain properties characterize pi, and especially its properties as a > member of the set of all real numbers. In different models of the real > numbers, pi might be a set of rational numbers, an equivalence class of > sequences of rational numbers, or even a point on a line. Whether or > not these objects are identical is irrelevant. The point is that all > these models behave equivalently in the ways that are significant for > real numbers. What color is pi? It doesn't matter. > Well, at the risk of posting on low blood caffeine, I believe it does > matter. Again, this is a topic to which I simply cannot direct my full > attention at present. What appears to be happening is a crossing of levels > of abstraction. This is how Russell tried to climb out of his paradox. My > instincts tell me that there is a contradiction to be found by reviewing > carefully all the statements and assumptions. Or perhaps there is a > necessary step missing which, if completed, would result in a > contradiction. Carefully reviewing to check for contradictions, eh? Why did nobody ever think to try doing that before? I suppose they didn't have your intuition to guide them. -- mike. === Subject: Re: Does the Calculus rest on Euclid? > Carefully reviewing to check for contradictions, eh? Why did nobody > ever think to try doing that before? I suppose they didn't have your > intuition to guide them. It is not the purpose of this work to cover the firm rock upon which the house of analysis is founded with a fake wooden structure of formalism - a structure that can fool the reader and, ultimately, the author into believing that it is the true foundation. Rather I shall show that this house is to a large degree build on sand - Hermann Weyl -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? > Carefully reviewing to check for contradictions, eh? Why did nobody > ever think to try doing that before? I suppose they didn't have your > intuition to guide them. It is not the purpose of this work to cover the firm rock upon which the > house of analysis is founded with a fake wooden structure of formalism - a > structure that can fool the reader and, ultimately, the author into > believing that it is the true foundation. Rather I shall show that this > house is to a large degree build on sand - Hermann Weyl A good deal of concrete has been intermixed with that sand since Weyl's time. That was over 50 years ago. === Subject: Re: Does the Calculus rest on Euclid? <8ddl829t0pji2ecj4apokibb9gvgjdpflm@4ax.com> <18OdnfOK3akkPBDZnZ2dnUVZ_v6dnZ2d@speakeasy.net I could see no way of going from the notion of an n-sided polygonal > approximation to the circle (or arc thereof) without invoking the Powers > on High. What are the Powers on High? If I say I'm going to compute the circumference of a circle by adding up the lengths of the sides of a regular n-gon, where n is a nonstandard integer, and then taking the standard part of the result am I invoking those powers in an especially virulent form, or am I avoiding invoking them at all? === Subject: Re: Does the Calculus rest on Euclid? >> I could see no way of going from the notion of an n-sided polygonal >> approximation to the circle (or arc thereof) without invoking the >> Powers on High. > What are the Powers on High? That is a very old question to which I will not attempt to provide an answer. > If I say I'm going to compute the > circumference of a circle by adding up the lengths of the sides of a > regular n-gon, where n is a nonstandard integer, and then taking the > standard part of the result am I invoking those powers in an especially > virulent form, or am I avoiding invoking them at all? I do not know what you mean by non-standard integer. -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? >> I could see no way of going from the notion of an n-sided polygonal >> approximation to the circle (or arc thereof) without invoking the >> Powers on High. What are the Powers on High? > That is a very old question to which I will not attempt to provide an > answer. Which seems to indicate that Hatto as no answer . > If I say I'm going to compute the > circumference of a circle by adding up the lengths of the sides of a > regular n-gon, where n is a nonstandard integer, and then taking the > standard part of the result am I invoking those powers in an especially > virulent form, or am I avoiding invoking them at all? > I do not know what you mean by non-standard integer. It is a number in field of non-standard reals. http://mathworld.wolfram.com/NonstandardAnalysis.html http://en.wikipedia.org/wiki/Nonstandard_analysis http://www.pupress.princeton.edu/titles/5812.html === Subject: Re: Does the Calculus rest on Euclid? <8ddl829t0pji2ecj4apokibb9gvgjdpflm@4ax.com> <18OdnfOK3akkPBDZnZ2dnUVZ_v6dnZ2d@speakeasy.net> That is a very old question to which I will not attempt to provide an > answer. That is very rude of you. If you had no intention of letting people know what you meant by the phase, if anything, you should not have used it. > If I say I'm going to compute the > circumference of a circle by adding up the lengths of the sides of a > regular n-gon, where n is a nonstandard integer, and then taking the > standard part of the result am I invoking those powers in an especially > virulent form, or am I avoiding invoking them at all? > I do not know what you mean by non-standard integer. And due to your discourtesy, I am under no obligation to tell you. === Subject: Re: Does the Calculus rest on Euclid? > If a real number is a set of rationals, and there is only one real >> number designated by the concept indicated by the phrase a real number, >> then, in the case of pi either the set called pi is identical to the >> object called pi, or we are avoiding (or more correctly, evading) the >> original topic. It doesn't make sense to talk about what the object called pi is. > Numbers don't exist in the same sense as the pen sitting on my desk. > Certain properties characterize pi, and especially its properties as a > member of the set of all real numbers. In different models of the real > numbers, pi might be a set of rational numbers, an equivalence class of > sequences of rational numbers, or even a point on a line. Whether or > not these objects are identical is irrelevant. The point is that all > these models behave equivalently in the ways that are significant for > real numbers. What color is pi? It doesn't matter. > Well, at the risk of posting on low blood caffeine, I believe it does > matter. If one has a number of models of a complete Archimedean ordered field, which are isomorphic as complete Archimedean ordered fields, what possible difference can it make to the properties of a complete Archimedean ordered field which one one takes as basic? > Again, this is a topic to which I simply cannot direct my full > attention at present. What appears to be happening is a crossing of levels > of abstraction. What in the world does that mean? > This is how Russell tried to climb out of his paradox. That is not how I read Russell's theory of types. Besides, there are much better ways now of avoiding Russell's paradox than that theory of types. > My > instincts tell me that there is a contradiction to be found by reviewing > carefully all the statements and assumptions. This NG is loaded with postings by those whose instincts tell them that. So far, none of them have been able to make a valid case that such contradictions exist. So that makes you only a minor member of a large and generally foolish group. > Or perhaps there is a > necessary step missing which, if completed, would result in a > contradiction. > I am confident that at least one concept has been accepted from the > properties I ascribe to the continuum in the development of the real > numbers. That is the concept of completed infinity. What is your concept of completed infinity, and why should anyone expect it to give rise to any contradictions? > This really wasn't what I had in mind when I started this thread. I was > really trying to get a handle on a rather elementary idea in geometry. > That concept is the idea of measuring the length of the arc subtending a > radial angle in a unit circle. I may have been well advised not to have > mentioned the calculus. > I could see no way of going from the notion of an n-sided polygonal > approximation to the circle (or arc thereof) without invoking the Powers > on High. Since the only satisfactory way of doing this so far involves the same sort of limiting process that defines definite integrals, how is one to avoid calculus in defining arc lenght? > I was thinking in terms of Euclid, Archimedes work on approximating pi is much closer to the mark. > but my reasoning was following > the line typically used in arriving at the arclength of a curve. Since > virtually all developments I have seen for the arclength theorem invoke > geometric analogs, its hard for me to identify the which is purely > analytical. In Analytical Geometry, everything geometrical has been made analytical. === Subject: Re: Does the Calculus rest on Euclid? >> Well, at the risk of posting on low blood caffeine, I believe it does >> matter. > If one has a number of models of a complete Archimedean ordered field, > which are isomorphic as complete Archimedean ordered fields, what > possible difference can it make to the properties of a complete > Archimedean ordered field which one one takes as basic? I'm trying to build up my understanding of the formal vocabulary used to discuss this topic so that the next time I return to it as a primary focus, at least the vocabulary will not be an obstacle to understanding. That task has to be left as a background job for now. >> Again, this is a topic to which I simply cannot direct my full >> attention at present. What appears to be happening is a crossing of >> levels of abstraction. > What in the world does that mean? Perhaps 'abstraction' wasn't the best term, but this is what I meant. Numbers are, to me, primitive 'objects'. They are points in 1-dimensional space. I say that numbers live at a primary level of abstraction. The concept of a set of primitive objects is, to me, 'more abstract'. what this really means is that the concept of a set references (refers to) primitive objects. If we define the meaning of 'set' so that sets only hold primitive objects, then there is no such thing as a set of sets. But now we want to talk about some aggregation of sets. (That which we typically call a set of sets). So we introduce the concept of a meta-set. Meta-sets can only contain sets. So we shall need meta-meta-sets, ad infinitum. If you don't proceed in that way then you can try to avoid Russell's paradox by defining away the set of all sets, while still allowing for sets of sets. But that begs the question of what to call the set of all sets not excluded by the set of all sets proclamation. >> My >> instincts tell me that there is a contradiction to be found by reviewing >> carefully all the statements and assumptions. > This NG is loaded with postings by those whose instincts tell them that. > So far, none of them have been able to make a valid case that such > contradictions exist. So that makes you only a minor member of a large > and generally foolish group. I am a fool-of-fools. I also have Weyl's skepticism as a motivation for my own. If Hermann Weyl did not accept the developments of Dedekind and Cauchy, I have good reason to remain skeptical. But I acknowledge that appeal to authority as is vacuous as appeal to majority when it comes to logical proof. I have not looked into what he presented in his later book _Algebraic Theory of Numbers_. He may well have become a True Believer ... but I doubt it. > What is your concept of completed infinity, and why should anyone > expect it to give rise to any contradictions? It's not so much that it would give rise to contradictions. It's more that it has the potential of being used to prove itself in a different guise. > Since the only satisfactory way of doing this so far involves the same > sort of limiting process that defines definite integrals, how is one to > avoid calculus in defining arc lenght? Well, that was part of the problem. I was trying to work from first principles without assuming The Calculus. I don't know when the concept of a radian was first introduced, but I believe the ancients had a notion of measuring arclength in terms of angle. >> I was thinking in terms of Euclid, > Archimedes work on approximating pi is much closer to the mark. >> but my reasoning was following >> the line typically used in arriving at the arclength of a curve. Since >> virtually all developments I have seen for the arclength theorem invoke >> geometric analogs, its hard for me to identify the which is purely >> analytical. > In Analytical Geometry, everything geometrical has been made analytical. [1] OK, so we define the perimeter of the unit circle as the set of all ordered pairs of real numbers {x,y} such that x^2+y^2 = 1. *) Is that set continuous? *) WTH does continuous mean in this context? *) How shall we express the notion of angle as it exists in Euclidian geometry in terms of Analytical Geometry? In Analytical Geometry of the most primitive form, there is no notion of a metric. I simply talk about multivalued functions over possibly multivalued domains. [1]I actually prefer your style of capitalization, but I don't believe it adheres to established convention. For example, I believe Special Relativity or Physics should be capitalized, but I don't belive that is common practice. -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? >> Well, at the risk of posting on low blood caffeine, I believe it does >> matter. If one has a number of models of a complete Archimedean ordered field, > which are isomorphic as complete Archimedean ordered fields, what > possible difference can it make to the properties of a complete > Archimedean ordered field which one one takes as basic? > I'm trying to build up my understanding of the formal vocabulary used to > discuss this topic so that the next time I return to it as a primary focus, > at least the vocabulary will not be an obstacle to understanding. That > task has to be left as a background job for now. >> Again, this is a topic to which I simply cannot direct my full >> attention at present. What appears to be happening is a crossing of >> levels of abstraction. > > What in the world does that mean? > Perhaps 'abstraction' wasn't the best term, but this is what I meant. > Numbers are, to me, primitive 'objects'. They are points in 1-dimensional > space. I say that numbers live at a primary level of abstraction. The > concept of a set of primitive objects is, to me, 'more abstract'. what > this really means is that the concept of a set references (refers to) > primitive objects. If we define the meaning of 'set' so that sets only > hold primitive objects, then there is no such thing as a set of sets. But that is not required. In fact there are systems in which sets can only hold other sets. The von Neumann-Bernays-Godel system, at least as far as the set of naturals naturals goes, is such a system. Setting such artificial restrictions as not letting a set be a member of a set destroys your position. > But > now we want to talk about some aggregation of sets. (That which we > typically call a set of sets). So we introduce the concept of a meta-set. Merely a non-empty set in most set theories. > Meta-sets can only contain sets. So we shall need meta-meta-sets, ad > infinitum. Since your meta-sets are merely ordinary sets in most current set theories, and anything like Russell's theory of types is no longer in use, your about a century out of date. > If you don't proceed in that way then you can try to avoid Russell's paradox > by defining away the set of all sets, while still allowing for sets of > sets. A much more common cure is to prohibit sets from being members of themselves. > But that begs the question of what to call the set of all sets not > excluded by the set of all sets proclamation. One cure for this is to require certain types of totalities to be proper classes which are not sets. http://mathworld.wolfram.com/SetClass.html >> My >> instincts tell me that there is a contradiction to be found by reviewing >> carefully all the statements and assumptions. Your careful review is not likely to find what thousands of other careful reviews have not found. This NG is loaded with postings by those whose instincts tell them that. > So far, none of them have been able to make a valid case that such > contradictions exist. So that makes you only a minor member of a large > and generally foolish group. > I am a fool-of-fools. I also have Weyl's skepticism as a motivation for my > own. If Hermann Weyl did not accept the developments of Dedekind and > Cauchy, I have good reason to remain skeptical. But I acknowledge that > appeal to authority as is vacuous as appeal to majority when it comes to > logical proof. I have not looked into what he presented in his later book > _Algebraic Theory of Numbers_. He may well have become a True Believer ... > but I doubt it. > What is your concept of completed infinity, and why should anyone > expect it to give rise to any contradictions? > It's not so much that it would give rise to contradictions. It's more that > it has the potential of being used to prove itself in a different guise. That doesn't tell us anything about what it is under any guise, so avoids answering the question. > Since the only satisfactory way of doing this so far involves the same > sort of limiting process that defines definite integrals, how is one to > avoid calculus in defining arc lenght? > Well, that was part of the problem. I was trying to work from first > principles without assuming The Calculus. I don't know when the concept of > a radian was first introduced, but I believe the ancients had a notion of > measuring arclength in terms of angle. For what kind of arcs? Circular arcs, perhaps, as being proportional to the central angles for fixed radii. But present day 'arcs' can be quite different. The general definition now being: An arc is a continuous image of a real interval. > I was thinking in terms of Euclid, Archimedes work on approximating pi is much closer to the mark. >> but my reasoning was following >> the line typically used in arriving at the arclength of a curve. Since >> virtually all developments I have seen for the arclength theorem invoke >> geometric analogs, its hard for me to identify the which is purely >> analytical. In Analytical Geometry, everything geometrical has been made analytical. > [1] > OK, so we define the perimeter of the unit circle as the set of all ordered > pairs of real numbers {x,y} such that x^2+y^2 = 1. That is one possible definition. It could also be defined as the image of the real interval {x: 0 ,+ x < 2*pi} under the mapping x -> (cos(x), sin(x)) > *) Is that set continuous? > *) WTH does continuous mean in this context? One definition that would allow one to call the graph of x^2+y^2 = 1 a continuous curve is noting that at each point of the graph either x is locally a continuous function of y or vice versa (and note that at all but finitely many points, each is locally a continuous function of the other). > *) How shall we express the notion of angle as it exists in Euclidian > geometry in terms of Analytical Geometry? In Analytical Geometry of the > most primitive form, there is no notion of a metric. I don't know what form of analytical geometry you are familiar with, but every one that I am familiar with is also called Cartesian geometry, in honor of its inventor, Rene de Carte, and comes with the built in Cartesian metric: the distance in the plane between points A = (a1,a2) and B = ( b1,b2) is Dist(A,B) = sqrt((a1-b1)^2+(a2-b2)^2), with similar metrics in higher dimensional spaces. > I simply talk about > multivalued functions over possibly multivalued domains. Such as ??? === Subject: Re: Does the Calculus rest on Euclid? <8ddl829t0pji2ecj4apokibb9gvgjdpflm@4ax.com> <18OdnfOK3akkPBDZnZ2dnUVZ_v6dnZ2d@speakeasy.net> infinitum. > Since your meta-sets are merely ordinary sets in most current set > theories, and anything like Russell's theory of types is no longer in > use, your about a century out of date. The ramfied theory of types isn't very popular, to say the least. But NFU+extensions is a good set theory alternative so far as I, not a set theorist, can determine, and it sort of has types. The assumptions of NFU are so weak you can prove it to be consistent, which seems like a grand way to start, and then you can add choice and infinity, and you are up and running. === Subject: Re: Does the Calculus rest on Euclid? On Sat, 10 Jun 2006 18:35:05 -0400, Hatto von Aquitanien >> On Fri, 09 Jun 2006 11:27:19 -0400, Hatto von Aquitanien >> A real number is a set S of rationals with the following properties: >> (i) S is non-empty >> (ii) S does not contain every rational >> (iii) If s, t are rational, s is in S and t > s, then t is in S >> (iv) S has no smallest element. >Pi is not a member of the set of rationals. >> No, it's not. The definition above does not require that >> every real number be rational. A real number _is_ a certain >> _set_ of rationals. > I do not accept your definition >of the real numbers. >> Guffaw. It's not my definition, it's a version of one of >> the standard definitions. Saying you do not accept >> a _definition_ is silly. >> Yes, by the definition above there _is_ a real number >> commonly called pi. And it's irrational. >If a real number is a set of rationals, and there is only one real number >designated by the concept indicated by the phrase a real number, then, in >the case of pi either the set called pi is identical to the object called >pi, Huh? Yes, the set called pi is identical to the object called pi. People have already explained exactly what set pi is. Unfortunately the explanations mentioned pi. Giving an actual definition of pi, starting from the definition of real number above, would take a lot of space. So let's talk about a different irrational, that's easier to define: sqrt(2) (the square root of 2). In fact sqrt(2) is equal to the set of all positive rationals r such that r^2 > 2. (r-squared is greater than 2). That's a perfectly well-defined set of rational numbers. And it is equal to the square root of two - when you define the product of two reals and then multipliy that number by itself you get 2. > or we are avoiding (or more correctly, evading) the original topic. >> No assumptions of continuity in sight. >To say that you can order things implies a concept of a continuum. >> What? >> Given the definition above, we _define_ x < y for real numbers >> x, y by saying x < y if and only if y is a proper subset of x. >But you cannot communicate any of that with out some appeal to my a priori >sense of continuum. Giggle. The fact that I cannot communicate that to _you_ proves very little. You're free to insist that your misconceptions are correct, even though you say that you haven't studied any of this, and even though it sounds like most of what you know about math comes via computer science and physics (which btw is not to suggest that they're bad mathematicians, just that they have no reason to worry about the sort of issue that we're discussing here, constructing R from the integers.) If you ever change your mind and decide you'd like to understand how the math actually works stop by again and ask - people will be happy to explain. > The very simple fact that I see these symbols ordered >on my computer screen involves my a priori concept of continuum. >implication, those of axiomatic set theory which is isomorphic to >symbolic logic. >> Uh no, set theory is symbolic logic _plus_ some axioms. >Which can be stated in terms of boolean algebra. >> Uh, no. >> Don't you get the feeling that just _maybe_ you should _learn_ >> something about these things, before proclaiming that mathemtaticians >> are all wrong? >Perhaps 'boolean algebra' may not be the best term to use. The distinctions >between 'boolean algebra', formal logic, propositional logic, etc. are not >always clearly maintained in computer science literature, and I would need >to carefully review the literature to draw clear distinctions. >What I am saying is that set theory begins with concepts which are identical >to those found expressed in terms of the primary logical connectives. >http://baldur.globalsymmetry.com/open-source/org/sth/math/logic.xhtml I understood that that's what you were saying. I understood what you meant by Boolean algebra - I wasn't quibbling about the distinction between Boolean algebra and propositional logic, if you want to take a broad view and say those are the same thing that's fine with me. Set theory involves much more than Boolean algebra. >It's really quite simple. If your system is logically consistent and >sufficiently powerful, then it provides a means of transitioning from any >given state (collection of statements forming an aggregate statement) to >another state defined by the rules for manipulating symbols. Any such >system can be codified in binary. >What is the geometry of a locally tangent space? ************************ David C. Ullrich === Subject: Re: Does the Calculus rest on Euclid? > If a real number is a set of rationals, and there is only one real number > designated by the concept indicated by the phrase a real number, then, in > the case of pi either the set called pi is identical to the object called > pi, or we are avoiding (or more correctly, evading) the original topic. There are several models of the real number field as a complete Archimedean ordered field. They can all be shown to be isomorphic as complete Archimedean ordered fields, so the issue of what their members are is irrelevant. An equally valid model starts with the subset of functions from the naturals to the rationals which are Cauchy sequences. These form a commutative ring under the pointwise sum and product definitions in which the set of such function which converge (in the rationals) to zero is a maximal ideal, so that the quotient ring mod that ideal if a field, and can be shown to be isomorphic as a complete Archimedean ordered field to the field of Dedekind cuts. > No assumptions of continuity in sight. >>To say that you can order things implies a concept of a continuum. What? Given the definition above, we _define_ x < y for real numbers > x, y by saying x < y if and only if y is a proper subset of x. > But you cannot communicate any of that with out some appeal to my a priori > sense of continuum. Where does the Cauchy sequence construction outlined above appeal to any notion of any continuum? > The very simple fact that I see these symbols ordered > on my computer screen involves my a priori concept of continuum. On my screen only discrete symbols (characters) appear. Even pictures are made up of discrete pixels on my screen. Where is this alleged continuum? === Subject: Re: Does the Calculus rest on Euclid? of reals, defined in terms of Dedekind cuts? > Every discussion I have seen of this topic requires a concept of completed > infinity, and a notion of infinite repeatability. What do you mean by completed infinity? Do you accept the existence of the set of positive integers? If not, I can understand why the reals are a problem. What do you mean by infinite repeatability? Are you an amateur constructivist? === Subject: Re: Does the Calculus rest on Euclid? of the set of positive integers? If not, I can understand why the reals > are a problem. Everyone who seems to believe you don't have an infinite set also seems to accept that there are classes of things which, if you completed them, would be infinite sets. In other words, they are quite happy to classify things as integers, and so forth. Now a question arises--why isn't that good enough for Dedekind cuts? What does moaning about completed infinities actually mean in practice? If I think the class of integers, by which I mean things satisfying Integer(x) then what am I not going to do that the fellow who thinks {x | Integer(x)} is going to do, and does this make a difference in this connection? === Subject: Re: Does the Calculus rest on Euclid? > I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I > have never felt satisfactorily convinced that the transition from the > Riemann sum approximation to a smooth curve is logically founded upon > axioms I have assumed at the outset. These axioms are those of formal > logic and those of Euclid. When using a geometric argument to justify the > transition from the chord-length approximation to a continuous curve - for > example in the typical proof of the arclength theorem - there seems to be > an unacknowledged step of faith. That is not a geometric argument. It is an intuition pump. One can do all of that without a scintilla of geomtry. All one needs are the non-neative integers. God made the integers. Man made all the rest. Bob Kolker === Subject: Re: integral of multiplied by sin(ax) or cos(ax) >I have the following problem: >Let f be a function which diverges to 0 on (0, infty). >Prove that int_0^infty f(x)sin(ax)dx > 0 This is not true. Instead of telling you what you meant to say, I think it's a better idea to say that _you_ should go back and read the problem more carefully. If you can't learn to read the problem correctly you're not going to have much chance of solving it. >and i'm interested what condition we can put on f in order to have also that >int_0^infty f(x)cos(ax)dx > 0. >What considers the first problem with sinus i think the following works: >Divide [0, infty) = union I_n, where >I_n = [pi*n/a,pi*(n+1)/a] so that sin(ax) has a constant sign on each of the I_n's and >A_n = int_I_n f(x)sin(ax)dx = (-1)^(n-1) int_I_n f(x)|sin(ax)|dx >and since f(x) diverges then |A_n| > |A_{n+1}|, so we have a series |A_1|-|A_2|+|A_3|-... > 0 >Is it ok? >And what can you say about my another question with cos. >I think the convexity would be enough, but i'm not sure. ************************ David C. Ullrich === Subject: Re: integral of multiplied by sin(ax) or cos(ax) >I have the following problem: >Let f be a function which diverges to 0 on (0, infty). What do you mean? A function can converge to 0, or diverge to infty, but I've never heard of a function that diverges to 0. >Prove that int_0^infty f(x)sin(ax)dx > 0 and i'm interested what >condition we can put on f in order to have also that >int_0^infty f(x)cos(ax)dx > 0. Perhaps you mean f(x) decreasing with f(x) -> 0 as x -> infty. Then it's true that if a > 0, int_{2 n pi/a}^{2 (n+1) pi/a} f(x) sin(ax) dx = int_{2 n pi/a}^{(2 n + 1) pi/a} (f(x) - f(x + pi/a)) sin(ax) dx > 0 and thus int_0^infty f(x) sin(ax) dx either converges to a positive number or diverges to +infty. If f is convex then for 0 <= x <= pi/(2 a) we have f(pi/a - x) + f(pi/a + x) <= f(x) + f(2 pi/a - x) so that int_{2 n pi/a}^{2 (n+1) pi/a} f(x) cos(ax) dx >= 0 and thus int_0^infty f(x) cos(ax) dx either converges to a nonnegative number or diverges (to +infty if f(x) -> 0 as x -> infty). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Chebychev inequality generalisation >Here is the Chebychev integral inequality: >int f(x)h(x)dx * int g(x)h(x)dx <= (>=) int h(x)dx * int f(x)g(x)h(x)dx, >where h(x) is a weigh function which is positive on the whole integration domain and f(x) and g(x) are both increasing(decreasing) integrable functions. Let's talk about one inequality at a time. You seem to be saying that (*) int f(x)h(x)dx * int g(x)h(x)dx <= int h(x)dx * int f(x)g(x)h(x)dx whenever f and g are both increasing. If so then (*) also holds whenever f and g are both decreasing. And then together with your other inequality it follows that we actually have int f(x)h(x)dx * int g(x)h(x)dx = int h(x)dx * int f(x)g(x)h(x)dx whenever f and g are both increasing. This seems unlikely... This time I _don't_ know what you actually meant to say. >I'm interested in whether there are some kind of generalisations of this inequality ,say, for many functions, not just two, or maybe a generalisation with generalise not number of functions but something else. >Any replies are welcome ************************ David C. Ullrich === Subject: Re: Chebychev inequality generalisation > On Fri, 09 Jun 2006 09:07:57 EDT, eugene >Here is the Chebychev integral inequality: >int f(x)h(x)dx * int g(x)h(x)dx <= (>=) int h(x)dx * > int f(x)g(x)h(x)dx, >where h(x) is a weigh function which is positive on > the whole integration domain and f(x) and g(x) are > both increasing(decreasing) integrable functions. > Let's talk about one inequality at a time. You seem > to be saying that > (*) int f(x)h(x)dx * int g(x)h(x)dx <= int h(x)dx * > int f(x)g(x)h(x)dx > whenever f and g are both increasing. If so then (*) > also holds > whenever f and g are both decreasing. > And then together with your other inequality it > follows that we > actually have > int f(x)h(x)dx * int g(x)h(x)dx = int h(x)dx * int > f(x)g(x)h(x)dx > whenever f and g are both increasing. This seems > unlikely... > This time I _don't_ know what you actually meant to > say. I wanted to say that if both of the function are of the same growth, i mean if they both are increasing or decreasing then the sing of the inequality is <=. If, say, one of them is increasing and the other one is decreasing then the sign of the inequality is >=. (To prove it just consider the double integral int int (f(x)-f(y))(g(x)-g(y))h(x)h(y)dxdy). Hope it is more clear now. === Subject: Solving 1 + p(1- p(x)) = g^[2](2 -x) , p known Have a look , p is a real strictly increasing p: R -> R known function can we find all g solutions? Alain === Subject: Re: Solving 1 + p(1- p(x)) = g^[2](2 -x) , p known alainverghote@yahoo.fr nous a r.8ecemment amicalement signifi.8e : > Have a look , > p is a real strictly increasing p: R -> R > known function > can we find all g solutions? Except the trivial solution g(x) = 1 + p(2-x), it's rather hard, depending on p. We have in fact : gog(x) = 1 + p(1 - p(2-x)) which is strictly increasing. And the problem is the classical probleme of finding solutions of gog=h where h is strictly increasing. The complexity depends on the number of points where h crosses y=x (in our case at least once) : in each interval or R in which h(x) is > x or < x, then we can find all solutions. -- Patrick === Subject: Re: Solving 1 + p(1- p(x)) = g^[2](2 -x) , p known reply-type=response Patrick Coilland nous a r.8ecemment amicalement signifi.8e : > alainverghote@yahoo.fr nous a r.8ecemment amicalement signifi.8e : >> Have a look , >> p is a real strictly increasing p: R -> R >> known function >> can we find all g solutions? > Except the trivial solution g(x) = 1 + p(2-x), it's rather hard, > depending on p. > We have in fact : gog(x) = 1 + p(1 - p(2-x)) which is strictly > increasing. > And the problem is the classical probleme of finding solutions of > gog=h where h is strictly increasing. > The complexity depends on the number of points where h crosses y=x (in > our case at least once) : in each interval or R in which h(x) is > x > or < x, then we can find all solutions. It's possible to change variable and write g(x) = 1 + h(2-x) hbeing the variable. Then we have to solve : h(1-h(x)) = p(1-p(x)) If we change again and say q(x) = p(1-x) strictly decreasing and h(x) = f(1-x) The problem becomes : Given q(x) from R in R, strictly decreasing, Find all functions f(x) defined in R such that : f(f(x)) = q(q(x)) The trivial solution is f = q but we have an infinity of solutions in the general case. -- Patrick === Subject: Re: Solving 1 + p(1- p(x)) = g^[2](2 -x) , p known <448ac1c9$0$943$ba4acef3@news.orange.fr> <448acf85$0$877$ba4acef3@news.orange.fr> Patrick Coilland a .8ecrit : > Patrick Coilland nous a r.8ecemment amicalement signifi.8e : > alainverghote@yahoo.fr nous a r.8ecemment amicalement signifi.8e : >> Have a look , >> p is a real strictly increasing p: R -> R >> known function >> can we find all g solutions? > Except the trivial solution g(x) = 1 + p(2-x), it's rather hard, > depending on p. > We have in fact : gog(x) = 1 + p(1 - p(2-x)) which is strictly > increasing. > And the problem is the classical probleme of finding solutions of > gog=h where h is strictly increasing. > The complexity depends on the number of points where h crosses y=x (in > our case at least once) : in each interval or R in which h(x) is > x > or < x, then we can find all solutions. > It's possible to change variable and write g(x) = 1 + h(2-x) hbeing the > variable. > Then we have to solve : > h(1-h(x)) = p(1-p(x)) > If we change again and say q(x) = p(1-x) strictly decreasing > and h(x) = f(1-x) > The problem becomes : > Given q(x) from R in R, strictly decreasing, > Find all functions f(x) defined in R such that : > f(f(x)) = q(q(x)) > The trivial solution is f = q but we have an infinity of solutions in > the general case. > -- > Patrick Always so very quick ! the trivial solution g(x) = 1 + p(2-x) , some words about the way you found it , please . The changing q(x) = p(1-x) and h(x) = f(1-x) seems nice and efficient , we have an infinity of solutions in the general case. f(f(x)) = q(q(x)) ?? Alain === Subject: Re: Solving 1 + p(1- p(x)) = g^[2](2 -x) , p known alainverghote@yahoo.fr nous a r.8ecemment amicalement signifi.8e : > we have an infinity of solutions in the general case. > f(f(x)) = q(q(x)) ?? I don't understand your question marks. We, and other posters, had a lot of exchanges with you on the fact that, in most cases, we have an infinity of solutions of fof = h, with h strictly increasing, if you have no constraint on f. If you add C-n constraint, we generally have an infinity of solutions. If you add C-inf constraint, I'm not sure. If you add analytic constraint, I don't know anything. -- Patrick === Subject: Re: Solving 1 + p(1- p(x)) = g^[2](2 -x) , p known <448ac1c9$0$943$ba4acef3@news.orange.fr> <448acf85$0$877$ba4acef3@news.orange.fr> <448adf47$0$836$ba4acef3@news.orange.fr> Patrick Coilland a .8ecrit : > alainverghote@yahoo.fr nous a r.8ecemment amicalement signifi.8e : > we have an infinity of solutions in the general case. > f(f(x)) = q(q(x)) ?? > I don't understand your question marks. > We, and other posters, had a lot of exchanges with you on the fact that, > in most cases, we have an infinity of solutions of fof = h, with h > strictly increasing, if you have no constraint on f. > If you add C-n constraint, we generally have an infinity of solutions. > If you add C-inf constraint, I'm not sure. > If you add analytic constraint, I don't know anything. > -- > Patrick OK Patrick , about the fact that we generally have an infinity of solutions. There is an other way to deal with h(1-h(x)) = p(1 - p(x)) 1 - h(1-h(x)) = 1 - p(1 - p(x)) or (1-h(x)) ^[2] = (1-p(x)) ^[2] so h(x) = p(x) works and for some values of (1-p(x)) ^[2] we may obtain other values of h , Alain === Subject: AN NWO EVENING OF SATANIC IGNORANCE: PROOF THAT THE WORLD HATES A CHRIST AN NWO EVENING OF SATANIC IGNORANCE: PROOF THAT THE WORLD HATES A CHRIST Raymond Ronald Karzcewski© -- A Living Christ A MIDNIGHT MEDICAL EMERGENCY dealt this Living Christ a fateful hand when He (I) was FORCIBLY EVICTED from a doctor's office at a hospital. After experiencing an evening of UNPRECEDENTED FLATULENCE which could have only been caused by Government/Media Sponsored/Supported Disinformation/Mind Control Agents using equipment manufactured by WEALTHY GOVERNMENT CONTRACTORS for a NWO takeover by the CIA/DHS/FBI/Military/Industrial Complex working for their Illuminati/Reptilian/Shapeshifter bosses, my rectum started to bleed. This isn't the first time that Government/Media Clandestine Operations have cause this Living Christ to experience volcanic flatulence and rectal bleeding, a little over a year ago I was ejected from an NWO sponsored diner for excessive flatulence that OFFENDED SATANIC The evening began as normally as any other evening at the Karzcewski household, my wife resting peacefully after dinner and I was partaking in dialogues with other Christs on the Internet when my bowels suddenly started rumbling in a violent, spastic manner. I experienced an almost nonstop explosive eruption of flatulence that started around 11 PM and did not taper off for nearly an hour. The CIA/NWO thugs have been hard at work upgrading the equipment which causes this flatulence and many Stealth Trucks may be heard coming and going filled with FEDERAL MILITIA TROOPS OPERATING ON CIVILIAN SOIL!!! I decided to demand treatment at the hospital when my wife started coughing and saying that the smell was almost unbearable. After mirroring her thusly I then left for the hospital and was promptly stopped by SATANIC SHERIFF'S DEPUTIES operating a DUALISTIC SPEED TRAP. After reminding them that SOVEREIGN LIVING CHRISTS do not need driver's licenses to operate noncommercial conveyances while traveling on private business, they noticed my bloodsoaked pants and escorted me to the hospital. For some reason these SATANIC COWARDS disappeared once I entered the facility and high-tailed it back to their master's depot to find wimpier prospects from which to pirate booty. They were part of the conspiracy to initiate the night's flatulence attack so that they would have opportunity to stop this Living Christ from making a divine journey. I was then escorted into an examination room where a very rude intern tried to interrogate me as to my sexual activities, implying that I had been inserting a phallic object into my rectum for sexual the depth of my extensive knowledge of all things, most especially the Government/Media Sponsored/Supported Disinformation/Mind Control Agents who assigned him to my case. Being the SATANIC COWARD that he was he slank back from whence he came to fetch a real doctor to examine my rectal bleeding. After waiting for an hour I decided to relieve my bladder and found that I was locked into the examination room. I then decided to urinate into the garbage can and was using the closet for privacy when a female nurse came into the room, saw what I was doing, and immediately screamed for security. She INSISTED that I was masturbating in the closet aand that I had inserted a handfull of tongue depressors into my rectum for sexual purposes. I attempted to explain that the tongue depressors were being used to stop my rectal bleeding but then a sudden volcanic eruption of gas blew them out on the floor and all over the closet door along with an emense blood spatter. In such a supposedly professional environment these medical experts were suddenly laughing at my embarrassing predicament, then pretending to choke on the smell of my flatulence. It is their SOLEMN DUTY to pretend not to smell my flatulence since it is a medical condition caused by the SATANIC EQUIPMENT buried in the hills behind my home. Such insolence in the presence of this Living, Divine, Sovereign Christ. They will soon pay, in this life or the next. After my PIPSQUEAKS I will rise and assume my throne as a LIVING GOD over all THIS EARTH WILL LEARN THE WRATH OF A LIVING CHRIST!!! Tremble in fear for a terrible fate awaits!!! Please feel free to redistribute this warning of doom to as many newsgroups as you see fit. The truth must be freely given far and wide if we are to take back this country and place the criminals hiding behind elected office behind bars where they belong. Until SATANIC Hell, isn't it? Raymond Ronald Karczewski© -- A Living Christ ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ In the illusion based world of man's thought, there exist healthy egos and unhealthy egos --visit my home page and meet the unhealthy ego's worst nightmare! http://www.arkenterprises.com June 10, 2006: DAY 495 The NATIONAL CONSUMER BOYCOTT giving Americans the opportunity to take back control over their country and their lives through participation. See: http://www.arkenterprises.com/dialch121.html#BOYCOTT YOUR ACTIONS COME FROM THE HEART, NOT your doubting SATANIC INTELLECT, and indeed, The Meek Shall Inherit the Earth. rk: P.S.: Attach the following SIGNATURE to all of your Internet Messaging. Let every person you know understand what is available to him/her in this historical NONVIOLENT REVOLUTION. rk: The Voting Booth HASN'T DONE IT. The Jury Box WON'T DO IT. Don't BET YOUR LIFE on the CARTRIDGE BOX either. rk: DO IT NONVIOLENTLY WITH: BOYCOTT, BOYCOTT, BOYCOTT!! TAKE BACK OUR COUNTRY -- STOP THE TALK, NOW WALK THE WALK! NATIONAL CONSUMER BOYCOTT TAKE-BACK BEGINS FEBRUARY 1, 2005 Spread The Word Throughout America! IT'S NOW OR NEVER!! http://www.arkenterprises.com/dialch121.html#BOYCOTT *********************************************************** BOYCOTT, BOYCOTT, BOYCOTT!! TAKE BACK OUR COUNTRY -- STOP THE TALK, NOW WALK THE WALK NATIONAL CONSUMER BOYCOTT TAKE-BACK BEGINS FEBRUARY 1, 2005 Spread The Word Throughout America! IT'S NOW OR NEVER!! http://www.arkenterprises.com/dialch121.html#BOYCOTT http://www.sherricorrell.com/2.htm ************************************************************ === Subject: Re: AN NWO EVENING OF SATANIC IGNORANCE: PROOF THAT THE WORLD HATES A CHRIST > After experiencing an evening of UNPRECEDENTED FLATULENCE BEANO - available just about anywhere. No harmful cameras or nano-bots in the pills. === Subject: Re: AN NWO EVENING OF SATANIC IGNORANCE: PROOF THAT THE WORLD HATES A CHRIST >> After experiencing an evening of UNPRECEDENTED FLATULENCE > BEANO - available just about anywhere. No harmful cameras or nano-bots > in > the pills. Well, there are, actually, but we don't want him to know. Oops! Did I just spill the ... um, beans? No pun intended, of course. ;) === Subject: Re: AN NWO EVENING OF SATANIC IGNORANCE: PROOF THAT THE WORLD HATES A CHRIST > After experiencing an evening of UNPRECEDENTED FLATULENCE >> BEANO - available just about anywhere. No harmful cameras or >> nano-bots in >> the pills. > Well, there are, actually, but we don't want him to know. > Oops! Did I just spill the ... um, beans? Well, that's blown it! We were hoping to get the Edward Scissor Hands bot in there. Now we'll have to fall back on the mosquitos. And they're so hard to train. Not to mention we lose so many. Sigh! > No pun intended, of course. ;) === Subject: Random Variable?? For each outcome e, X(e) can take multiple values. I am considering if X is valid Random Variable? === Subject: Re: Random Variable?? > For each outcome e, X(e) can take multiple values. I am considering if > X is valid Random Variable? For each value e on the first die, the total X of the two dice can take multiple values. Is that what you mean? === Subject: Re: Random Variable?? >> For each outcome e, X(e) can take multiple values. I am considering if >> X is valid Random Variable? > For each value e on the first die, the total X of the two dice can > take multiple values. Is that what you mean? He's talking about a multivalued function. AFAIK it doesn't make any sense in the definition of a random variable but maybe there is a way to generalize it. i.e. if X is a random variable then the meaning of X is defined as P(X(w) = x) = P({w | X(w) = x|) if X is multivalued then I'm not sure how one could rectify it. maybe X(w) = x could be generalized to x in X(w). Not sure if that makes sense though. I'm not sure how one would end up with a multivalued random variable though... === Subject: Correlation Median and Mean value? Hoping I got the right news group for this... I have recorded data with nearly exactly the same median and mean value. Example: max value 61.5 min value 61 median 61.3 mean 61.303 It shows a very small difference between both values (app. 1/1000 in the range between max/min). Question: Does there exist some statistical meaning if both values are identical, what does it mean for my measurements? Pierre === Subject: Re: Correlation Median and Mean value? > Hoping I got the right news group for this... > I have recorded data with nearly exactly the same median and mean value. > Example: > max value 61.5 > min value 61 > median 61.3 > mean 61.303 > It shows a very small difference between both values (app. 1/1000 in the > range between max/min). > Question: Does there exist some statistical meaning if both values are > identical, what does it mean for my measurements? It means that the distribution of the data is symmetrical about the mean value. === Subject: Re: Correlation Median and Mean value? > Hoping I got the right news group for this... > I have recorded data with nearly exactly the same median and mean value. > Example: > max value 61.5 > min value 61 > median 61.3 > mean 61.303 > It shows a very small difference between both values (app. 1/1000 in the > range between max/min). > Question: Does there exist some statistical meaning if both values are > identical, what does it mean for my measurements? > It means that the distribution of the data is symmetrical about the > mean value. Nope. For example, the median and mean of {1,4,5,7,8} are both 5, but the distribution is not symmetrical about the mean. === Subject: Re: Correlation Median and Mean value? > Hoping I got the right news group for this... > > I have recorded data with nearly exactly the same median and mean value. > > Example: > max value 61.5 > min value 61 > median 61.3 > mean 61.303 > > It shows a very small difference between both values (app. 1/1000 in the > range between max/min). > > Question: Does there exist some statistical meaning if both values are > identical, what does it mean for my measurements? > It means that the distribution of the data is symmetrical about the > mean value. > Nope. For example, the median and mean of {1,4,5,7,8} are both 5, but > the distribution is not symmetrical about the mean. If the distribution is symmetric then the mean equals the median, but not necessarily conversely. === Subject: Clearity always nadulent Mail-To-News-Contact: abuse@dizum.com === Subject: Re: Anybody got an IQ over 20 here? Such a silly, silly person! He must divide his like into compartments because otherwise he could no function. In a previous generation his type claimed that the Moon Landings were faked. === Subject: Re: An ancient sudoku <87pshqqnl8.fsf@nonospaz.fatphil.org> The identity of the 75 invisible units: http://web.gvdnet.dk/GVD002393/castle.gif === Subject: Re: anybody got an IQ over 200 here? Discussion, linux) <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> <8n7j825tl7drtoradjk4cj3o93d6r3n2dj@4ax.com Especially for a guy who only travelled outside the US twice in his > life, even while his father was the US envoy to China, the US > ambassador to the US, the Director of the CIA, a congressman, the > Vice President of the US and even the President of the US. And one > of those trips was just to see a donkey show in Tijuana. I don't think this is true. It may well be true that Bush didn't travel as much as one expects for a pampered rich kid or as much as is prudent for a US president, but I think the claim he traveled outside the US only twice is just false. George was not exactly what you would call well-traveled. Campaign staffers claimed that he had taken more than a dozen trips outside the U.S., although they admitted that the vague figure included many, many trips to Mexico and Canada. Bush made a month-long excursion to China while his father was stationed there, which the New York Times summed up as trying to date Chinese women (unsuccessfully) during a visit to Beijing in 1975. He had visited Israel and Egypt with the National Governors Association, and also the African country of Gambia. Later on in the campaign, Bush staffers claimed that he has also visited England, Scotland, and Italy, as well as vacationed in France and Bermuda. This was not very impressive to the people of Europe, who have to cross international borders just to take their kids to Legoland. (Rotten.com may not be the best citation, but they are certainly not biased in favor of Bush.) There are so, so many really, really, really good reasons to regret Bush's presidency and to question why he was elected twice. Might as well stick to the truth. [Note followups] -- Damn John Jay. Damn everyone who won't damn John Jay. Damn everyone who won't put lights in his windows and sit up all night damning John Jay. -- Political graffiti from late 18th c. Boston === Subject: Re: anybody got an IQ over 200 here? <4ecrl8F1efu6iU1@individual.net> <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> The list was off the internet so how accurate it is I don't know. But >> George W. Bush's job is much more difficult -- especially in foreign >> affairs -- than people realize. A lot is at stake. How do you know that? === Subject: Re: anybody got an IQ over 200 here? <4ecrl8F1efu6iU1@individual.net> <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> The list was off the internet so how accurate it is I don't know. But >> George W. Bush's job is much more difficult -- especially in foreign >> affairs -- than people realize. A lot is at stake. > How do you know that? Foreign affairs involves the very survival of the United States. Right now -- and since WWII -- the U.S. has been the king of the Hill. Politics being what it is one nation after another has attempted to knock the U.S. and it's allies down from the top. Serbia was one such nation. China would like to knock the U.S. down, but is being careful. And, Russia sits on the sidelines waiting to see who falls and then will take advantage of the situation if both competitors have diminished their strength in the competition. And, the French just sit there building their economy by not participating in anything. They are abiding by the theory that resting in a safe position will make them stronger when the chips fall the wrong way. What a world, and the President has to deal with all of this and even more, because there are many nations doing 'this and that' that have not been mentioned here. The Earth is a pretty rough battleground. Right now the U.S. is at peak strength both militarily and economically, a real credit to the man in the Office of the Presidency. And, right now, that man is George W. Bush. tomcat === Subject: Re: anybody got an IQ over 200 here? > The list was off the internet so how accurate it is I don't know. But > George W. Bush's job is much more difficult -- especially in foreign > affairs -- than people realize. A lot is at stake. >> How do you know that? >Foreign affairs involves the very survival of the United States. Right >now -- and since WWII -- the U.S. has been the king of the Hill. >Politics being what it is one nation after another has attempted to >knock the U.S. and it's allies down from the top. Serbia was one such >nation. >China would like to knock the U.S. down, but is being careful. And, >Russia sits on the sidelines waiting to see who falls and then will >take advantage of the situation if both competitors have diminished >their strength in the competition. >And, the French just sit there building their economy by not >participating in anything. They are abiding by the theory that resting >in a safe position will make them stronger when the chips fall the >wrong way. >What a world, and the President has to deal with all of this and even >more, because there are many nations doing 'this and that' that have >not been mentioned here. The Earth is a pretty rough battleground. >Right now the U.S. is at peak strength both militarily and >economically, a real credit to the man in the Office of the Presidency. > And, right now, that man is George W. Bush. Interesting. And what were you saying in the year, say, 1999? === Subject: Re: anybody got an IQ over 200 here? <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> This is critical - and would make for better discussions, so long as the logic is not self-contradictory and the facts can be reasonably established! The PC stuff muddies the waters. -- foolsrushin.' > David, > Please don't muddle political correctness with logic or facts! > You'll throw the PC crowd into a tailspin. > Like this one better? > http://www.truthorfiction.com/rumors/p/presidentialiq.htm > 147 Franklin D. Roosevelt (D) > 132 Harry Truman (D) > 122 Dwight D. Eisenhower (R) > 174 John F. Kennedy (D) > 126 Lyndon B. Johnson (D) > 155 Richard M. Nixon (R) > 121 Gerald Ford (R) > 175 James E. Carter (D) > 105 Ronald Reagan (R) > 099 George HW Bush (R) > 182 William J. Clinton (D) > 091 George W. Bush (R) > The non-partisan researchers who evaluated the twelve presidents > Non-partisan. It was a panel of college professors, right? > President G. W. Bush was rated the lowest of all the Republicans with > an IQ of 91. The six Democrat presidents had IQs with an average of 156, > Bet no one saw that coming. > Bush's IQs were challenged, but I don't recall whether the complaint > was that they were estimated too high or too low. > Har. Neat trick, graduating from Yale with a below-average IQ. > Or even meeting the entrance requirements. > Dubya made a 1206 on his SAT (566v+640m, before they > dumbed down the scale in 1994), which roughly translates to > an IQ score of 123 (http://www.gnxp.com/MT2/archives/002360.html) > to 129 (http://members.shaw.ca/delajara/Pre1974SAT.html). > Clinton's supposed IQ of 182 is 5.5 standard deviations above the > norm, meaning he would be expected to have easily made a perfect > 1600 on his SAT score. He reputedly has an actual IQ closer to 135, > which puts him somewhere around a 1320~1350 SAT score. === Subject: Re: anybody got an IQ over 200 here? On 9 Jun 2006 10:45:53 -0700, Reef Fish >> http://www.aceviper.net/aceviper_net/ace_intelligence/aceviper_famous_people _ iq_list/aceviper_famous_people_iq_list.html >> Question: What do Hillary Clinton, Madonna, and Adolf Hitler have in > common? >> Answer: The same IQ >> Correction: Hitler's IQ was 1 point higher. > This list of I.Q.'s clearly shows that George W. Bush's I.Q. is on a > par with Abraham Lincoln's and Andrew Jackson's. It is, moreover, > clearly superior to George Washington's, John F. Kennedy's, and Ulysses > S. Grant's. >> But that was a list laundered by the White House of the current > administration. :-) >> George W. Bush's IQ is comparable to Andy Warhol's except Bush never > produced a 25 hour movie of the same scene in Manhattan. Bush has > staged a 25 month theatre of the absurd in Iraq though. >> -- Bob. >>The list was off the internet so how accurate it is I don't know. But >>George W. Bush's job is much more difficult -- especially in foreign >>affairs -- than people realize. A lot is at stake. >> Especially for a guy who only travelled outside the US twice in his >> life, even while his father was the US envoy to China, the US >> ambassador to the US, the Director of the CIA, a congressman, the Vice >> President of the US and even the President of the US. And one of >> those trips was just to see a donkey show in Tijuana. >Is that the only thing you know about George W. Bush? LOL! No, unfortunately, it isn't. I also know that he was drunk until he was 40 (at least), he ran every business venture and official government venture he was involved in into the ground (except the Texas Rangers, where he was just a name used by the people who actually ran the business to attract money and influence) and he's the only oil man in Texas who couldn't find oil in Texas. > For a more >detailed account of the real Mr. Bush, see: >http://www.rotten.com/library/bio/presidents/george-w-bush/ Figgers you'd be a Bush supporter. One of the top 29%, eh? LOL! >> It's amazing to me that the US voters put such an incurious and >> massively ignorant fellow in charge of something as difficult as >> foreign affairs. >Why is it amazing to YOU, Wilbur Slice? I find it much more >amazing that someone like yourself, who is supposed to have had >SOME education in the US, could be as ignorant as you are, >about IQ, about statistics, and about other areas, as your have >adequated exhibited (as proof) your ignorance. >You are truly in the BOTTOM .02% of the US population in IQ >and intelligence. LOL. Ooooo... looks like I struck a nerve. Now I have a stalker. LOL! >-- Bob. === Subject: Re: anybody got an IQ over 200 here? <8n7j825tl7drtoradjk4cj3o93d6r3n2dj@4ax.com> > Especially for a guy who only travelled outside the US twice in his >> life, even while his father was the US envoy to China, the US >> ambassador to the US, the Director of the CIA, a congressman, the Vice >> President of the US and even the President of the US. And one of >> those trips was just to see a donkey show in Tijuana. >Is that the only thing you know about George W. Bush? > LOL! No, unfortunately, it isn't. I also know that he was drunk > until he was 40 (at least), he ran every business venture and official > government venture he was involved in into the ground (except the > Texas Rangers, where he was just a name used by the people who > actually ran the business to attract money and influence) and he's the > only oil man in Texas who couldn't find oil in Texas. > For a more >detailed account of the real Mr. Bush, see: >http://www.rotten.com/library/bio/presidents/george-w-bush/ > Figgers you'd be a Bush supporter. Besides your proven ignorance about IQ, IQ tests, and statistics, you've added one more -- your ignorance about where I stand on Bush as well as your reading comprehension. Do you really think that piece was written by a Bush supporter? Read it again. I know it's hard for you hold your attention span more than 3 lines at a time, but still ... >> It's amazing to me that the US voters put such an incurious and >> massively ignorant fellow in charge of something as difficult as >> foreign affairs. >Why is it amazing to YOU, Wilbur Slice? I find it much more >amazing that someone like yourself, who is supposed to have had >SOME education in the US, could be as ignorant as you are, >about IQ, about statistics, and about other areas, as your have >adequated exhibited (as proof) your ignorance. >You are truly in the BOTTOM .02% of the US population in IQ >and intelligence. LOL. > Ooooo... looks like I struck a nerve. Now I have a stalker. LOL! You flatter yourself, as always. -- Bob. === Subject: Reed Solomon encoding I've a basic question about encoding with Reed Solomon code. Suppose that C is a RS-code of length n = q-1 over the field F_q, designed distance d and dimension k. Let b a primitive element in F_q. The generator of the code is g(x) = (x-b) (x-b^2)...(x-b^(d-1)). Now, if I want to encode an element p(x) in F_q [x] of degree less or equal to k-1, I must multiply g(x) p(x). The result is an element of C. Is it correct to encode in this way? (I suppose the answer is affermative). However I have read about a systematic encoding. This method seems different. And the result of encoding too. So, if a choose a fixed RS-code C, with a generator g(x), using the first method and the systematic encoding I have two different result. Can you confirm this? === Subject: Re: Reed Solomon encoding > I've a basic question about encoding with Reed Solomon code. Suppose > that C is a RS-code of length n = q-1 over the field F_q, designed > distance d and dimension k. Let b a primitive element in F_q. The > generator of the code is g(x) = (x-b) (x-b^2)...(x-b^(d-1)). Now, if I > want to encode an element p(x) in F_q [x] of degree less or equal to > k-1, I must multiply g(x) p(x). The result is an element of C. Is it > correct to encode in this way? (I suppose the answer is affermative). > However I have read about a systematic encoding. This method seems > different. And the result of encoding too. So, if a choose a fixed > RS-code C, with a generator g(x), using the first method and the > systematic encoding I have two different result. Can you confirm this? Yes. Different choice for a generator matrix results in a different encoding function. In linear algebra this means that when you change to a different basis, the coordinates (for the same vector) will change in a predictable way. Nothing mysterious about that. The systematic encoding (that really amounts to using a generator matrix reduced to a row echelon form + permutation of columns, if need be) is often used, because it has the practical advantage of easy recovery of the information symbols (= the coordinates). Also simple hardware for efficient computation of the check symbols exists in the case of RS-codes. However, this is only a choice. Upshots: 1) do it any way you want, but 2) make sure that the receivers and the transmitter are both doing it the same way. Jyrki === Subject: Re: spherical coordinates question > Hello > I want to know how to find the lat/lon coordinates of a rectangle, say > 100x300 km, placed at an arbitrary location somewhere at Earth's > mid-latitudes. The rectangle is placed at an angle, i.e., its sides do > not lie on great circles or meridians. The important thing is that its > sides remain exactly 100/300 km apart, along the respective sides of > the rectangle. > Any help much appreciated! > Evan It is entirely possible to construct a figure such as you describe, as far as the side lengths are concerned. It's easiest, of course, if the earth is allowed to be considered spherical. Having never actually worked it out, I have no idea as to the internal angles at the corners of the figure-- I suspect they won't be right angles. Your problem interests me, and I'll work on this later today. Grover Hughes === Subject: Re: spherical coordinates question > Hello > I want to know how to find the lat/lon coordinates of a rectangle, say > 100x300 km, placed at an arbitrary location somewhere at Earth's > mid-latitudes. The rectangle is placed at an angle, i.e., its sides do > not lie on great circles or meridians. The important thing is that its > sides remain exactly 100/300 km apart, along the respective sides of > the rectangle. > Any help much appreciated! > Evan It is impossible to construct a rectangle on the surface of the earth. Something has to give. It's not clear precisely what remain exactly 100/300 km apart means when the sides are curved lines lying on the surface of a sphere (or approximate sphere), or how one would define this measurement so as to ensure that the resulting rectangle-like shape is uniquely determined from some positional parameters. For an approximate answer you could take a plane rectangle, tangential to the earth at the rectangle's centre, and project the rectangle's edges, perpendicularly to the plane of the rectangle, onto the earth's surface. But this wouldn't satisfy the exactly 100/300 km apart requirement. === Subject: L^2 with a complex measure I have a complex measure absolutely continuous w.r.t. the Lebesgue measure on the half-line [0,+infty[ defined as follows: d m(t) = p(t) dt with p: [0,+infty[ --> C (say p(t) = p_1(t) + i p_2(t) with p_ j real functions). Can I define (namely HOW can I define) L^2(dm(t)) ? Since the measure is positive have I to take the absolute value of p(t) ? In other words: have I to say that some function f:[0,infty[ --> C is in L^2(dm(t)) if and only if { int_0^infty |f(t)|^2 d m(t) is a (finite) complex number } OR { have I to to consider int_0^infty |f(t)|^2 d |m(t)| } ?? I thought to decompose the measure in its real and imaginary parts (and eventually in its real positive, real negative, imaginary positive and imaginary negative parts) but I am not sure how to handle the four (eventually eight) cases that arise since I consider a space of COMPLEX functions and a COMPLEX measure. I hope someone can help me. Can someone also give me some references where I can find an explicit definition of L^2 in the case of complex measure ? === Subject: Re: L^2 with a complex measure >I have a complex measure absolutely continuous w.r.t. the Lebesgue >measure on the half-line [0,+infty[ defined as follows: >d m(t) = p(t) dt >with p: [0,+infty[ --> C >(say p(t) = p_1(t) + i p_2(t) with p_ j real functions). >Can I define (namely HOW can I define) L^2(dm(t)) ? >Since the measure is positive have I to take the absolute value of p(t) >In other words: >have I to say that some function >f:[0,infty[ --> C >is in L^2(dm(t)) if and only if >int_0^infty |f(t)|^2 d m(t) >is a (finite) complex number >have I to to consider >int_0^infty |f(t)|^2 d |m(t)| >I thought to decompose the measure in its real and imaginary parts (and >eventually in its real positive, real negative, imaginary positive and >imaginary negative parts) but I am not sure how to handle the four >(eventually eight) cases that arise since I consider a space of COMPLEX >functions and a COMPLEX measure. I am no analyst, but for a signed or complex measure m, wouldn't it make the most sense to say f in L^2 iff |integral(|f|^2 dm)| < infty? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: L^2 with a complex measure > I am no analyst, but for a signed or complex measure m, wouldn't it > make the most sense to say f in L^2 iff |integral(|f|^2 dm)| < infty? No. For example if m has positive and negative values, there could be cancellation in your integral. For the L^2 space, you want the analog of absolute convergence. There should be a positive measure (called |m|) so that dm(t) = h(t) d|m|(t) for some complex function h of absolute value 1. Then define L^2(m) to be L^2(|m|). === Subject: Re: L^2 with a complex measure >I have a complex measure absolutely continuous w.r.t. the Lebesgue >measure on the half-line [0,+infty[ defined as follows: >d m(t) = p(t) dt >with p: [0,+infty[ --> C >(say p(t) = p_1(t) + i p_2(t) with p_ j real functions). >Can I define (namely HOW can I define) L^2(dm(t)) ? >Since the measure is positive have I to take the absolute value of p(t) >In other words: >have I to say that some function >f:[0,infty[ --> C >is in L^2(dm(t)) if and only if >int_0^infty |f(t)|^2 d m(t) >is a (finite) complex number No. >have I to to consider >int_0^infty |f(t)|^2 d |m(t)| Ues, that would be the standard definition. >I thought to decompose the measure in its real and imaginary parts (and >eventually in its real positive, real negative, imaginary positive and >imaginary negative parts) but I am not sure how to handle the four >(eventually eight) cases that arise since I consider a space of COMPLEX >functions and a COMPLEX measure. >I hope someone can help me. >Can someone also give me some references where I can find an explicit >definition of L^2 in the case of complex measure ? ************************ David C. Ullrich === Subject: Re: L^2 with a complex measure I meant Since the measure is NOT positive... > I have a complex measure absolutely continuous w.r.t. the Lebesgue > measure on the half-line [0,+infty[ defined as follows: > d m(t) = p(t) dt > with p: [0,+infty[ --> C > (say p(t) = p_1(t) + i p_2(t) with p_ j real functions). > Can I define (namely HOW can I define) L^2(dm(t)) ? > Since the measure is positive have I to take the absolute value of p(t) > In other words: > have I to say that some function > f:[0,infty[ --> C > is in L^2(dm(t)) if and only if > int_0^infty |f(t)|^2 d m(t) > is a (finite) complex number > OR > have I to to consider > int_0^infty |f(t)|^2 d |m(t)| > ?? > I thought to decompose the measure in its real and imaginary parts (and > eventually in its real positive, real negative, imaginary positive and > imaginary negative parts) but I am not sure how to handle the four > (eventually eight) cases that arise since I consider a space of COMPLEX > functions and a COMPLEX measure. > I hope someone can help me. > Can someone also give me some references where I can find an explicit > definition of L^2 in the case of complex measure ? === Subject: Re: seperated-proper-complete > sjang schreef: >> l.s. >> I am confused about the following: >> With the Zariski topology, isn't every affine space compact? >> How does one show this? >> sjang > I didn't quite finish my question: > for the affine line A, any open covering {U_i} has a finite > subcovering: U_1 say, is the line without finitely many points, and > these can be covered by finitely many open sets U_j. > I think this can be generalized by looking at the coordinate ring and > use that it is noetherian. You are right by having a look at the coordinate ring. But the noetherian condition is not needed here, but the fact that if an ideal is the unit ideal then unity (1) can be represented as a *finite* sum of elements in the ideal. > But what is really confusing me is the following: a variety Y is said > to be complete if the map Y --> point is proper, id est, if the > projection YxZ-->Z is closed for every variety Z. > If we take Y=Z=A, the affine line, we see that A is not complete; > projecting the closed > V(xy-1) onto A gives A{0}, which is open. > However, in a Bourbaki textbook, I found the following: Y is compact if > and only if Y-->point is proper. > this seems to contradict, and I don't see where it goes wrong; probably > in the definitions. > If somebody can clarify?! What is confusing you is terminology differently used in various fields of mathematics - as this has been discussed here already many times. The property you are talking of above is quasi-compactness (any open covering has a finite sub-covering). Certainly any affine variety (or affine scheme in a more general algebro-geometric setting) is quasi-compact. In topology the term compact means, firstly, Hausdorff and, secondly, quasi-compact. Now, varieties and scheme are *not* Hausdorff in the Zariski topology, in general. So there is some need to have a nice equivalent for the property of a topological space to be compact. To make this new property work for objects in algebraic geometry (e.g. varieties or schemes), the property of completeness is defined as you did above in categorical terms in the light of the theorem you quoted from Bourbaki. Now, this might go back to one of Serre's fundamental theorem, namely GAGA (G.8eom.8etrie alg.8ebrique et g.8eom.8etrie analytique), as it links the notions 'complete' (in algebraic geometry) and 'compact' (in topology): If you have a (quasi-projective (*)) variety over the complex numbers, then it is complete iff it is compact in the associated strong topology. (*) Ignore this adjective at first reading as it is a technical assumption to make things work for Serre's famous theorem. In brief, it is of advantage to distinguish the notions 'quasi-compact', 'compact' and 'complete' from each other. I hope these thoughts can be helpful for you. Best wishes, J. === Subject: Re: book or online page on factoring... >Does anybody know of a good book or online page on various factoring >formulas. >For example: >how to factor x^n + 1 where n is odd, but more complicated cases as >well, and many special cases. Are you trying to factor an integer (e.g. of the form x^n + 1 where x and n are integers)? Or are you trying to factor the polynomial x^n + 1 for a fixed n ? (Any factorization of the latter type implies a factorization of the former type, but e.g. x^3 + 1 = (x+1) (x^2 - x + 1) is a factorization into two irreducibles in Z[x] while 5^3 + 1 = (5+1) (5^2 - 5 + 1) is a factorization in Z whose factors are not irreducible.) Factorization in Z[x] is an easier problem and there are algorithms for it. For the special case of factorizations of x^n + 1 with n odd, the complete factorization into irreducibles is x^n + 1 = - Prod Phi_d(-x) , the product taken over all divisors d of n, where Phi_d is the d-th cyclotomic polynomial. For example in the first factorization I gave above you can see the use of Phi_1(x) = x-1 and Phi_3(x) = x^2 + x + 1. I'm a little hard-pressed to figure out what you might want that's a more complicated case than cyclotomic polynomials without going into the complete question of how to factor general polynomials in Z[x]. dave === Subject: Complex exponential function? I am confused by the complex exponential! The complex exponential shows up in solutions to the Schrodinger wave equation, with the general form e^(i*a*t), where a is a real constant. I do not understand what this means! You can rewrite this according to Euler's formula as cos(a*t)-i*sin(a*t), but I do not understand how the time evolution of this function works. Should the imaginary component be thought of as a sort of extra dimension to the time function, and analyzed like a vector? -n. === Subject: Re: Complex exponential function? > I am confused by the complex exponential! > The complex exponential shows up in solutions to the Schrodinger wave > equation, with the general form e^(i*a*t), where a is a real constant. > I do not understand what this means! You can rewrite this according > to Euler's formula as cos(a*t)-i*sin(a*t), but I do not understand how > the time evolution of this function works. Should the imaginary > component be thought of as a sort of extra dimension to the time > function, and analyzed like a vector? > -n. First of all, Euler's formula says e^(i*a*t) = cos(a*t) + i * sin(a*t). It can be thought of as 2 real components, cos(a*t) & sin(a*t), certainly, but this likely misses something of how the factor interacts with other complex valued quantities. exp(i*a*t) is a complex number of absolute value (modulus) 1, and multiplication by this amount in the complex field has an effect of rotating the value multiplied around the origin through an angle a*t (in radians, counterclockwise). Therefore discussion of this in connection with, say, electronic circuitry or even solutions of Schroedinger's eqn., will often refer to it as a phase shaft. === Subject: A more intuitive and simple way of thinking about the Monty Hall problem Quick Summary of the problem from Wikipedia: >>A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. First you point toward a door, he says. Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door. You begin by pointing to door number 1. The host shows you that door number 3 has a goat.<< 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you picking a goat with your first pick. 2) Since it's been established that there is a 2/3 chance you will pick a goat, this means there's a 2/3 chance the host will be forced to pick the LAST REMAINING goat (thus leaving the car unpicked), since he cannot pick the car. 3) This means that there's a 2/3 chance that the remaining choice for you to switch to will be the car, since there's a 2/3 chance he was forced to pick the remaining goat, leaving the car free. I like my explanation better than all the others I've seen. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem Check out this link: http://mathforum.org/dr.math/faq/faq.monty.hall.html === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Quick Summary of the problem from Wikipedia: >>A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > I like my explanation better than all the others I've seen. An even simpler and more intuitive explanation is as follows. Imagine doing the experiment 3,000,000 times. Say in every experiment you choose door #1. Well, in 1,000,000 of the experiments, the car is behind door #1, so you should not switch (if you have x-ray vision), but in 2,000,000 of the experiments, the car is behind the other door that Monte did not choose. You can also use this type of representation to analyze a different situation. Suppose that, instead of the scenario you have described, Monte actually tosses a coin to decide on which of doors #2 or #3 to open. After doing this, he opens a door that happens to hide a goat. Now what are your win probabilities? Well, now you have a 1/2 chance of winning if you switch or don't switch. You can see this by counting up the different possibilities. R.G. Vickson === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Imagine doing the experiment 3,000,000 times. Say in every Then it doesn't really matter which strategy I follow. One million cars would already be too much for me, as a practical matter, and another million - would I even notice the difference? === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem Then it doesn't really matter which strategy I follow. One million > cars would already be too much for me, as a practical matter, and > another million - would I even notice the difference? You are missing the point. Of the 3 million experiments, you win by switching in 2 million of them. Therefore, the probability is 2/3 of winning if you switch. Of course, you can get this directly from standard probability calculations, but SOME people understand these matters better if they think about the problem in terms of doing a large number of experiments and looking at relative frequencies of outcomes. RGV === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem >> Imagine doing the experiment 3,000,000 times. Say in every >> Then it doesn't really matter which strategy I follow. One million >> cars would already be too much for me, as a practical matter, and >> another million - would I even notice the difference? > You are missing the point. Of the 3 million experiments, you win by > switching in 2 million of them. Therefore, the probability is 2/3 of > winning if you switch. Of course, you can get this directly from > standard probability calculations, but SOME people understand these > matters better if they think about the problem in terms of doing a > large number of experiments and looking at relative frequencies of > outcomes. No, I did not miss your point. It is, indeed, a helpful way to think about the probability, if you can imagine having millions of cars without thinking of all the problems they would bring. Not all is maths. Your argument should be convincing, but I think there is a participant in this thread who insists that switchers win at probability 1/2. Can you find a way to convince him? I don't think that's a mathematical problem, either. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem Imagine doing the experiment 3,000,000 times. Say in every >> Then it doesn't really matter which strategy I follow. One million >> cars would already be too much for me, as a practical matter, and >> another million - would I even notice the difference? > You are missing the point. Of the 3 million experiments, you win by > switching in 2 million of them. Therefore, the probability is 2/3 of > winning if you switch. Of course, you can get this directly from > standard probability calculations, but SOME people understand these > matters better if they think about the problem in terms of doing a > large number of experiments and looking at relative frequencies of > outcomes. > No, I did not miss your point. It is, indeed, a helpful way to think > about the probability, if you can imagine having millions of cars > without thinking of all the problems they would bring. Not all is > maths. > Your argument should be convincing, but I think there is a participant > in this thread who insists that switchers win at probability 1/2. Can > you find a way to convince him? I don't think that's a mathematical > problem, either. The probability of winning if you always switch is equivalent to the probability you initially picked a goatdoor = 2/3 The probability of winning if you always stick is equivalent to the probability you initially picked the cardoor = 1/3 Pretty convincing to me. Perhaps it helps to think of a different game where there are now 99 doors with goats and 1 door with car. You pick a door. The host open 98 doors with goats. He asks if you want to switch. Does anyone think you have only a 1/2 chance of winning if you switch, here? -- mike. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Imagine doing the experiment 3,000,000 times. Say in every >> Then it doesn't really matter which strategy I follow. One million >> cars would already be too much for me, as a practical matter, and >> another million - would I even notice the difference? > > You are missing the point. Of the 3 million experiments, you win by > switching in 2 million of them. Therefore, the probability is 2/3 of > winning if you switch. Of course, you can get this directly from > standard probability calculations, but SOME people understand these > matters better if they think about the problem in terms of doing a > large number of experiments and looking at relative frequencies of > outcomes. > No, I did not miss your point. It is, indeed, a helpful way to think > about the probability, if you can imagine having millions of cars > without thinking of all the problems they would bring. Not all is > maths. > Your argument should be convincing, but I think there is a participant > in this thread who insists that switchers win at probability 1/2. Can > you find a way to convince him? I don't think that's a mathematical > problem, either. > The probability of winning if you always switch is equivalent to the > probability you initially picked a goatdoor = 2/3 > The probability of winning if you always stick is equivalent to the > probability you initially picked the cardoor = 1/3 > Pretty convincing to me. > Perhaps it helps to think of a different game where there are now 99 > doors with goats and 1 door with car. You pick a door. The host open 98 > doors with goats. He asks if you want to switch. Does anyone think you > have only a 1/2 chance of winning if you switch, here? > -- > mike. A more intuitive and simple way of thinking about the Monty Hall problem The simplest way to view this problem is to just accept the fact that the goat is the better prize and choose it. You can at least get milk and meat from a goat. However, due to fossil fuel depletion the car will be good only as a flower planter in your backyard, with a side function of providing a radio to play so you can keep in touch with the world's future exploding complexity, assuming we survive. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > The simplest way to view this problem is to just accept the fact > that the goat is the better prize and choose it. You can at least > get milk and meat from a goat. However, due to fossil fuel > depletion the car will be good only as a flower planter in your > backyard, with a side function of providing a radio to play so you > can keep in touch with the world's future exploding complexity, > assuming we survive. Nice. Are we allowed to switch to the door that Monty opened? That would again make switching the better strategy. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem The a priori probability that the car is not behind the door that you picked is 2/3. This probability does not change since the revealation of the first goat is not a random event.. To any bridge players - Doesn't this sort of remind you of the Theory of Restricted Choice? Bill J. > Quick Summary of the problem from Wikipedia: >>A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > I like my explanation better than all the others I've seen. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Quick Summary of the problem from Wikipedia: >>A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > I like my explanation better than all the others I've seen. I find this one slightly simpler: 1) Pick a door. Since there are 2 goats and 1 car, there's a 2/3 chance that the car is behind one of the other doors. 2) There is still at least one goat behind the other doors. The host - knowing what is behind them - picks one with a goat. This does not bring any new information. 3) So there is still a 2/3 chance that the other unpicked door has the car. This is a nice applet: http://math.ucsd.edu/~crypto/Monty/monty.html Dirk Vdm === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem >A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. This is important. If the host doesn't say he's playing by these rules, the problem is actually unsolvable. (This has been thoroughly discussed here before, but _someone_ will undoubtedly start it up again. 8-) ) > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > I like my explanation better than all the others I've seen. > I find this one slightly simpler: > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. > 3) So there is still a 2/3 chance that the other unpicked > door has the car. Yes, that's what he said. While this idea won't hold up mathematically, it is the principle behind the solution. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > I find this one slightly simpler: > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. The host's revealing his choice tells the contestant something he did not know before. > 3) So there is still a 2/3 chance that the other unpicked > door has the car. > Yes, that's what he said. > While this idea won't hold up mathematically, it is the principle > behind the solution. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem I find this one slightly simpler: > > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. > The host's revealing his choice tells the contestant something he did > not know before. No matter which door you picked first the host always picks a door with a goat behind it to reveal. This provides no useable information. Using the switching strategy : Event 1) You pick the cardoor, the host reveals one of the goatdoors, you switch and lose. Event 2) You pick one of the goatdoors, the host reveals the other goatdoor, you switch and win. These events partition the sample space. The only randomness introduced is by your initial choice so Event 1) has probability 1/3 and leads to loss; Event 2) has probability 2/3 and leads to success. Asking is switching likely to result in a win? is equivalent to asking is it more likely you selected a goatdoor than a cardoor?. The probability of winning if you always switch is equivalent to the probability you initially picked a goatdoor = 2/3 The probability of winning if you always stick is equivalent to the probability you initially picked the cardoor = 1/3 -- mike. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > I find this one slightly simpler: > > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. > The host's revealing his choice tells the contestant something he did > not know before. > No matter which door you picked first the host always picks a door with > a goat behind it to reveal. This provides no useable information. Nonsense. It tells you that the main prize is not behind monty's door. That's information. And you can use it to deduce that there's now a 2/3 probability that the prize is behind the until now unchosen door. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem <878xo4hwm5.fsf@nonospaz.fatphil.org > I find this one slightly simpler: > > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. > > The host's revealing his choice tells the contestant something he did > not know before. > No matter which door you picked first the host always picks a door with > a goat behind it to reveal. This provides no useable information. > Nonsense. It tells you that the main prize is not behind > monty's door. That's information. And you can use it to > deduce that there's now a 2/3 probability that the prize > is behind the until now unchosen door. Except you already knew before Monty picked that he would be opening a door with a goat behind it. Knowing which of the two doors he picks provides no information as to which action you should take : switch to the remaining door or stick with your original choice. You already chose your door with a 1/3 chance of being right and already knew he was going to be opening a door with a goat behind it. -- mike. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > > I find this one slightly simpler: > > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. > > The host's revealing his choice tells the contestant something he did > not know before. > > No matter which door you picked first the host always picks a door with > a goat behind it to reveal. This provides no useable information. > Nonsense. It tells you that the main prize is not behind > monty's door. That's information. And you can use it to > deduce that there's now a 2/3 probability that the prize > is behind the until now unchosen door. > Except you already knew before Monty picked that he would be opening a > door with a goat behind it. Knowing which of the two doors he picks > provides no information as to which action you should take : switch to > the remaining door or stick with your original choice. You already > chose your door with a 1/3 chance of being right and already knew he > was going to be opening a door with a goat behind it. Just because the strategy is known, it doesn't mean the action is known. What you are saying seems equivalent to Kasparov's opening move also containing no information. The computer just needs to pull out the predetermined response from its opening book. Ditto Kasparov's second move, of course... Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem <878xo4hwm5.fsf@nonospaz.fatphil.org> <87ver7h9lh.fsf@nonospaz.fatphil.org > > I find this one slightly simpler: > > 1) Pick a door. Since there are 2 goats and 1 car, there's > a 2/3 chance that the car is behind one of the other doors. > > 2) There is still at least one goat behind the other doors. > The host - knowing what is behind them - picks one with > a goat. This does not bring any new information. > > The host's revealing his choice tells the contestant something he did > not know before. > > No matter which door you picked first the host always picks a door with > a goat behind it to reveal. This provides no useable information. > > Nonsense. It tells you that the main prize is not behind > monty's door. That's information. And you can use it to > deduce that there's now a 2/3 probability that the prize > is behind the until now unchosen door. > Except you already knew before Monty picked that he would be opening a > door with a goat behind it. Knowing which of the two doors he picks > provides no information as to which action you should take : switch to > the remaining door or stick with your original choice. You already > chose your door with a 1/3 chance of being right and already knew he > was going to be opening a door with a goat behind it. > Just because the strategy is known, it doesn't mean the action > is known. What you are saying seems equivalent to Kasparov's > opening move also containing no information. The computer > just needs to pull out the predetermined response from its > opening book. Ditto Kasparov's second move, of course... > Phil > -- Bizzare to compare a game of Chess to a game with a handful of possible states, no? Anyhow in this case not only is the strategy known, the action *is* known. Monty *is* going to open one of the doors and it *is* going to have a goat behind it. It doesn't matter which door it is. Once you make your initial random choice the game is completely deterministic (it doesn't matter which of the two goatdoors he selects if you happened to guess correctly). Which of the doors Monty picks to reveal doesn't provide any information pertinent to your best move that you didn't have before he made his choice. -- mike. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Just because the strategy is known, it doesn't mean the action > is known. What you are saying seems equivalent to Kasparov's > opening move also containing no information. The computer > just needs to pull out the predetermined response from its > opening book. Ditto Kasparov's second move, of course... > Bizzare to compare a game of Chess to a game with a handful of possible > states, no? No. One's finite, the other's finite. We're big boys here on sci.math, ten to the few doesn't phase us. [SNIP - missing the point of my previous post.] Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem <878xo4hwm5.fsf@nonospaz.fatphil.org> <87ver7h9lh.fsf@nonospaz.fatphil.org> <87ac8jh1z5.fsf@nonospaz.fatphil.org > Just because the strategy is known, it doesn't mean the action > is known. What you are saying seems equivalent to Kasparov's > opening move also containing no information. The computer > just needs to pull out the predetermined response from its > opening book. Ditto Kasparov's second move, of course... > Bizzare to compare a game of Chess to a game with a handful of possible > states, no? > No. One's finite, the other's finite. > We're big boys here on sci.math, ten to the few doesn't phase us. Relevence of the Chess comparison? Monty is not a free agent in the Monty Hall problem. He provides no information because he has no choice in his actions. Kasparov has 22 possible opening moves and it gets hairer from there on. > [SNIP - missing the point of my previous post.] > Phil Yeah, I have no idea what you were trying to say in your post or how it was at all relevent to the Monty Hall problem. Obviously a comprehension problem on my part... -- mike. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Anyhow in this case not only is the strategy known, the action *is* > known. Monty *is* going to open one of the doors and it *is* going to > have a goat behind it. It doesn't matter which door it is. Once you > make your initial random choice the game is completely deterministic > (it doesn't matter which of the two goatdoors he selects if you > happened to guess correctly). Which of the doors Monty picks to reveal > doesn't provide any information pertinent to your best move that you > didn't have before he made his choice. That's not so. You now know which door he picked. happens when we SWITCH. Switch * [GOAT] 1 * * * * [CAR] 1/3 * * * * * 0 * [CAR] * NOT Switch * * * * Switch * 1 * [CAR] * * 2/3 * * * * *[GOAT] * * NOT Switch* 0 * [GOAT] (or is plus; and is times) P(win/switching) = (1/3) (0) + (2/3)(1) = 2/3 happens when we do NOT SWITCH. Switch * [GOAT] 0 * * * * [CAR] 1/3 * * * * * 1 * [CAR] * NOT Switch * * * * Switch * 0 * [CAR] * * 2/3 * * * * *[GOAT] * * NOT Switch* 1 * [GOAT] P(win / not switching) = (1/3) (1) + (2/3)(0) = 1/3 === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem <878xo4hwm5.fsf@nonospaz.fatphil.org> <87ver7h9lh.fsf@nonospaz.fatphil.org> known. Monty *is* going to open one of the doors and it *is* going to > have a goat behind it. It doesn't matter which door it is. Once you > make your initial random choice the game is completely deterministic > (it doesn't matter which of the two goatdoors he selects if you > happened to guess correctly). Which of the doors Monty picks to reveal > doesn't provide any information pertinent to your best move that you > didn't have before he made his choice. > That's not so. You now know which door he picked. > happens when we SWITCH. > Switch * [GOAT] > 1 * > * > * > * [CAR] > 1/3 * * > * * > * 0 * [CAR] > * NOT Switch > * > * > * Switch > * 1 * [CAR] > * * > 2/3 * * > * * > *[GOAT] > * > * > NOT Switch* > 0 * [GOAT] > (or is plus; and is times) > P(win/switching) = (1/3) (0) + (2/3)(1) = 2/3 > happens when we do NOT SWITCH. > Switch * [GOAT] > 0 * > * > * > * [CAR] > 1/3 * * > * * > * 1 * [CAR] > * NOT Switch > * > * > * Switch > * 0 * [CAR] > * * > 2/3 * * > * * > *[GOAT] > * > * > NOT Switch* > 1 * [GOAT] > P(win / not switching) = (1/3) (1) + (2/3)(0) = 1/3 Uh, great diagrams, really. If I were struggling to multiply 2/3 by 1 I would now be much clearer. How does the above illustrate that it matters which door he picked? You don't seem to mention which door he picked so all I can conclude is that you think it doesn't matter either. I'm not sure exactly which part of my post your it's not so is referring to... -- mike. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem <878xo4hwm5.fsf@nonospaz.fatphil.org> <87ver7h9lh.fsf@nonospaz.fatphil.org> Anyhow in this case not only is the strategy known, the action *is* > known. Monty *is* going to open one of the doors and it *is* going to > have a goat behind it. It doesn't matter which door it is. Once you > make your initial random choice the game is completely deterministic > (it doesn't matter which of the two goatdoors he selects if you > happened to guess correctly). Which of the doors Monty picks to reveal > doesn't provide any information pertinent to your best move that you > didn't have before he made his choice. > That's not so. You now know which door he picked. > happens when we SWITCH. > Switch * [GOAT] > 1 * > * > * > * [CAR] > 1/3 * * > * * > * 0 * [CAR] > * NOT Switch > * > * > * > * Switch > * 1 * [CAR] > * * > 2/3 * * > * * > *[GOAT] > * > * > NOT Switch* > 0 * [GOAT] > (or is plus; and is times) > P(win/switching) = (1/3) (0) + (2/3)(1) = 2/3 > happens when we do NOT SWITCH. > Switch * [GOAT] > 0 * > * > * > * [CAR] > 1/3 * * > * * > * 1 * [CAR] > * NOT Switch > * > * > * > * Switch > * 0 * [CAR] > * * > 2/3 * * > * * > *[GOAT] > * > * > NOT Switch* > 1 * [GOAT] > P(win / not switching) = (1/3) (1) + (2/3)(0) = 1/3 > Uh, great diagrams, really. If I were struggling to multiply 2/3 by 1 I > would now be much clearer. How does the above illustrate that it > matters which door he picked? [...] If you don't know by now, there's no use explaining any further. You need to Google for probability trees (or take a course in probability), learn about how probability trees are used, re-read Silver's post, and only then post again. http://mathforum.org/dr.math/faq/faq.boy.girl.html is probably a good place to start online. --- Christopher Heckman === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem >> I find this one slightly simpler: >> 1) Pick a door. Since there are 2 goats and 1 car, there's >> a 2/3 chance that the car is behind one of the other doors. >> 2) There is still at least one goat behind the other doors. >> The host - knowing what is behind them - picks one with >> a goat. This does not bring any new information. In order not to give trolls like Virgil a chance, I should have said that it does not bring any new *relevant* information. >> The host's revealing his choice tells the contestant something he did >> not know before. >> No matter which door you picked first the host always picks a door with >> a goat behind it to reveal. This provides no useable information. > Nonsense. It tells you that the main prize is not behind > monty's door. That's information. And you can use it to > deduce that there's now a 2/3 probability that the prize > is behind the until now unchosen door. It does not bring new *relevant* information about - the car being behind either the already picked door, or - the car being behind one of the two unpicked doors. Both parties knew this, so it does not bring new information about whether you should switch or not. Dirk Vdm === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem Another simple way to look at it is, there is a 1/3 chance you were right the first time. This stays the same if you don't switch. Therefore, if you switch, the probability of success must be 2/3, since the total must be 1. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem <18693057.1149974392261.JavaMail.jakarta@nitrogen.mathforum.org>, > Another simple way to look at it is, there is a 1/3 chance you were right the > first time. This stays the same if you don't switch. Therefore, if you > switch, the probability of success must be 2/3, since the total must be 1. Wrong! it is the probability of success plus the probability of failure for any fixed strategy which must add to one. There is no reason why the probabilities of success for different strategies have to add up to anything in particular. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > <18693057.1149974392261.JavaMail.jakarta@nitrogen.math > forum.org>, > Another simple way to look at it is, there is a 1/3 > chance you were right the > first time. This stays the same if you don't > switch. Therefore, if you > switch, the probability of success must be 2/3, > since the total must be 1. > Wrong! it is the probability of success plus the > probability of failure > for any fixed strategy which must add to one. > There is no reason why the probabilities of success > for different > strategies have to add up to anything in particular. There are only two possibilities. The car is behind the door originally picked (prob =1/3) or behind the door remaining (prob = 2/3). These must add to 1. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem ><18693057.1149974392261.JavaMail.jakarta@nitrogen.mathforum.org>, >> Another simple way to look at it is, there is a 1/3 chance you were right the >> first time. This stays the same if you don't switch. Therefore, if you >> switch, the probability of success must be 2/3, since the total must be 1. > Wrong! it is the probability of success plus the probability of failure > for any fixed strategy which must add to one. > There is no reason why the probabilities of success for different > strategies have to add up to anything in particular. Except for the fact that in this game, you win by switching iff you lose by not switching. There are only two unopened doors, after all, and exactly one of them is a winner. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Quick Summary of the problem from Wikipedia: >>A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > I like my explanation better than all the others I've seen. But your analysis ignores the case in which your original pick is the door with the car. Since switching in that case is a mistake, your overall chance of winning the car will turn out to be 1/3 by not switching and 1/2 by switching. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > But your analysis ignores the case in which your original pick is the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. The sum of the probabilities of winning and not winning have have to add up to unity. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > But your analysis ignores the case in which your original pick is > the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. > The sum of the probabilities of winning and not winning have have to > add up to unity. It is the probability of winning plus the probability of losing for a fixed strategy which must add up to 1. The probability of winning is dependent on one's choice of strategy. The overall probabilities of winning are 1/3 if one always chooses not to switch or 1/2 if one always chooses to switch. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. > The sum of the probabilities of winning and not winning have have to > add up to unity. > It is the probability of winning plus the probability of losing for a > fixed strategy which must add up to 1. > The probability of winning is dependent on one's choice of strategy. > The overall probabilities of winning are 1/3 if one always chooses not > to switch or 1/2 if one always chooses to switch. That's just wrong. If you always switch, you win 2/3's of the time. Again, it has do add up to one. If you never switch, you win 1/3 and lose 2/3. If you always switch, then you win 2/3 and lose 1/3. If you switch evenly between stratergies, you win 1/2 and lose 1/2. The probability of winning plus the probability of losing always equals unity, if there are no other options. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. > The sum of the probabilities of winning and not winning have have to > add up to unity. > It is the probability of winning plus the probability of losing for a > fixed strategy which must add up to 1. > The probability of winning is dependent on one's choice of strategy. > The overall probabilities of winning are 1/3 if one always chooses not > to switch or 1/2 if one always chooses to switch. Can you post the code of the program you claim to have that proves this? === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem >A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > I like my explanation better than all the others I've seen. > But your analysis ignores the case in which your original pick is the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. I would say think of it this way: Saying that there is a 2/3 chance of the contestant initially picking a there's a 2/3 chance of the host picking the LAST REMAINING goat. This is so because, by construction, if the contestant picks a goat, the host must pick the last remaining goat, leaving the car free. Thus, if there's a 2/3 chance of the contestant picking a goat, then we're talking about also a 2/3 chance of the host picking the last remaining goat, leaving there being a 2/3 chance of there being a car in the remaining door. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem > Quick Summary of the problem from Wikipedia: >>A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > > I like my explanation better than all the others I've seen. > But your analysis ignores the case in which your original pick is the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. > I would say think of it this way: > Saying that there is a 2/3 chance of the contestant initially picking a > there's a 2/3 chance of the host picking the LAST REMAINING goat. This > is so because, by construction, if the contestant picks a goat, the > host must pick the last remaining goat, leaving the car free. Thus, if > there's a 2/3 chance of the contestant picking a goat, then we're > talking about also a 2/3 chance of the host picking the last remaining > goat, leaving there being a 2/3 chance of there being a car in the > remaining door. This may be intuitive and simple to you , but it is wrong according to several modelings , including by me, on computers, and these models have run long enough so that if the chance of winning by switching were actually 2/3, it would have been obvious. And it was obviously not. === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem Quick Summary of the problem from Wikipedia: >>A thoroughly honest > game-show host has placed a car behind one of three doors. There is a > goat behind each of the other doors. You have no prior knowledge that > allows you to distinguish among the doors. First you point toward a > door, he says. Then I'll open one of the other doors to reveal a > goat. After I've shown you the goat, you make your final choice whether > to stick with your initial choice of doors, or to switch to the > remaining door. You win whatever is behind the door. > You begin by pointing to door number 1. The host shows you that door > number 3 has a goat.<< > > 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you > picking a goat with your first pick. > > 2) Since it's been established that there is a 2/3 chance you will > pick a goat, this means there's a 2/3 chance the host will be forced to > pick the LAST REMAINING goat (thus leaving the car unpicked), since he > cannot pick the car. > > 3) This means that there's a 2/3 chance that the remaining choice for > you to switch to will be the car, since there's a 2/3 chance he was > forced to pick the remaining goat, leaving the car free. > > I like my explanation better than all the others I've seen. > > But your analysis ignores the case in which your original pick is the > door with the car. Since switching in that case is a mistake, your > overall chance of winning the car will turn out to be 1/3 by not > switching and 1/2 by switching. No, it doesn't. That case is not ignored because the case is considered when I say that the probability is picking a goat initially is 2/3. The car is included in the three possiblities. Thus, it is not ignored. > I would say think of it this way: > Saying that there is a 2/3 chance of the contestant initially picking a > there's a 2/3 chance of the host picking the LAST REMAINING goat. This > is so because, by construction, if the contestant picks a goat, the > host must pick the last remaining goat, leaving the car free. Thus, if > there's a 2/3 chance of the contestant picking a goat, then we're > talking about also a 2/3 chance of the host picking the last remaining > goat, leaving there being a 2/3 chance of there being a car in the > remaining door. > This may be intuitive and simple to you , but it is wrong according to > several modelings , including by me, on computers, and these models have > run long enough so that if the chance of winning by switching were > actually 2/3, it would have been obvious. And it was obviously not. Play with this experiment if you wish. http://math.ucsd.edu/~crypto/Monty/monty.html === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem >> Quick Summary of the problem from Wikipedia: >>A thoroughly honest >> game-show host has placed a car behind one of three doors. There is a >> goat behind each of the other doors. You have no prior knowledge that >> allows you to distinguish among the doors. First you point toward a >> door, he says. Then I'll open one of the other doors to reveal a >> goat. After I've shown you the goat, you make your final choice whether >> to stick with your initial choice of doors, or to switch to the >> remaining door. You win whatever is behind the door. >> You begin by pointing to door number 1. The host shows you that door >> number 3 has a goat.<< >> 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you >> picking a goat with your first pick. >> 2) Since it's been established that there is a 2/3 chance you will >> pick a goat, this means there's a 2/3 chance the host will be forced to >> pick the LAST REMAINING goat (thus leaving the car unpicked), since he >> cannot pick the car. >> 3) This means that there's a 2/3 chance that the remaining choice for >> you to switch to will be the car, since there's a 2/3 chance he was >> forced to pick the remaining goat, leaving the car free. >> I like my explanation better than all the others I've seen. >> But your analysis ignores the case in which your original pick is the >> door with the car. Since switching in that case is a mistake, your >> overall chance of winning the car will turn out to be 1/3 by not >> switching and 1/2 by switching. >> I would say think of it this way: >> Saying that there is a 2/3 chance of the contestant initially picking a >> there's a 2/3 chance of the host picking the LAST REMAINING goat. This >> is so because, by construction, if the contestant picks a goat, the >> host must pick the last remaining goat, leaving the car free. Thus, if >> there's a 2/3 chance of the contestant picking a goat, then we're >> talking about also a 2/3 chance of the host picking the last remaining >> goat, leaving there being a 2/3 chance of there being a car in the >> remaining door. > This may be intuitive and simple to you , but it is wrong according to > several modelings , including by me, on computers, and these models have > run long enough so that if the chance of winning by switching were > actually 2/3, it would have been obvious. And it was obviously not. So sci.math has *two* crackpots? Dirk Vdm === Subject: Re: A more intuitive and simple way of thinking about the Monty Hall problem Quick Summary of the problem from Wikipedia: >>A thoroughly honest >> game-show host has placed a car behind one of three doors. There is a >> goat behind each of the other doors. You have no prior knowledge that >> allows you to distinguish among the doors. First you point toward a >> door, he says. Then I'll open one of the other doors to reveal a >> goat. After I've shown you the goat, you make your final choice whether >> to stick with your initial choice of doors, or to switch to the >> remaining door. You win whatever is behind the door. >> You begin by pointing to door number 1. The host shows you that door >> number 3 has a goat.<< >> 1) Since there are 2 goats and 1 car, there's a 2/3 chance of you >> picking a goat with your first pick. >> 2) Since it's been established that there is a 2/3 chance you will >> pick a goat, this means there's a 2/3 chance the host will be forced to >> pick the LAST REMAINING goat (thus leaving the car unpicked), since he >> cannot pick the car. >> 3) This means that there's a 2/3 chance that the remaining choice for >> you to switch to will be the car, since there's a 2/3 chance he was >> forced to pick the remaining goat, leaving the car free. >> I like my explanation better than all the others I've seen. >> But your analysis ignores the case in which your original pick is the >> door with the car. Since switching in that case is a mistake, your >> overall chance of winning the car will turn out to be 1/3 by not >> switching and 1/2 by switching. >> I would say think of it this way: >> Saying that there is a 2/3 chance of the contestant initially picking a >> there's a 2/3 chance of the host picking the LAST REMAINING goat. This >> is so because, by construction, if the contestant picks a goat, the >> host must pick the last remaining goat, leaving the car free. Thus, if >> there's a 2/3 chance of the contestant picking a goat, then we're >> talking about also a 2/3 chance of the host picking the last remaining >> goat, leaving there being a 2/3 chance of there being a car in the >> remaining door. > This may be intuitive and simple to you , but it is wrong according to > several modelings , including by me, on computers, and these models have > run long enough so that if the chance of winning by switching were > actually 2/3, it would have been obvious. And it was obviously not. > So sci.math has *two* crackpots? Maybe he's just a bad programmer. > Dirk Vdm === Subject: Re: JSH: SF might work, bad luck? [jstevh@msn.com] >> [...] >> Try it. Rotate a parabola and see what you get. >> If it seems too abstract, post what you get. >> Like, rotate a parabola 45 degrees. >> Post the result. [David C. Ullrich] > Ok. If you rotate a parbola 45 degrees the result > is a parabola. [Tim Peters] >> There you go, fooling yourself with those pure math ideas again. Like >> he said, _try_ it. It's amazing! It turns into a hyperbola, until you >> rotate your _head_ 45 degrees too. Then it looks like a parabola again, >> unless you rotate your head the wrong way, in which case it vanishes >> entirely. In between, keep your head still and it turns into a square >> at 1, 4, 9, 16, 25 and 36 degrees, a circle at around 31.4 degrees, and >> if you keep going you get a really cool pointillist sketch of what looks >> like the the Four Horsemen of the Apocalypse at 66 3/5 degrees. Applied >> math rulz! [David C. Ullrich] > Wow - I'm surprised to see you saying these Things here. I mean > I know you think you have protection, but you must realize > that the NSA is listening in - you really think your guys are > bigger than them? LOL -- what makes you imagine my guys _aren't_ them? Ask yourself who stands to gain the most from an unsuspected ability to crack RSA in microseconds. That's right, the Russian mafia. Then ask yourself who the CIA really works for. Then ask yourself who the Russian mafia really works for. Put it all together and you get exponentially growing piles of rabbits. > Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, ... degr So you knew that all along but were playing possum. Clever! What you don't know is James's proof that distinct Fibonacci numbers are never congruent modulo 360. You may even believe it's trivial to prove the contrary. Here's the correct proof: suppose F_i =/= F_j but F_i = F_j modulo 360. Then rotating a parabola F_i degrees gives the same number of rabbits as rotating a parabola F_j degrees, and rotating a parabola F_i - F_j degrees gives F_i - F_j rabbits. But F_i =/= F_j, so F_i - F_j is not 0. Therefore F_i - F_j may as well be 45. But if you rotate a parabola F_i - F_j = 45 degrees, you get a hyperbola, and a hyperbola is not 45 rabbits. QED. > You never tire of finding _new_ ways to make a public fool > of yourself. Well, I tried, but I'm afraid that argument was too clear === Subject: Re: JSH: SF might work, bad luck? On Sat, 10 Jun 2006 14:53:53 -0400, Tim Peters [...] >> Try it. Rotate a parabola and see what you get. >> If it seems too abstract, post what you get. >> Like, rotate a parabola 45 degrees. >> Post the result. >[David C. Ullrich] >> Ok. If you rotate a parbola 45 degrees the result >> is a parabola. >[Tim Peters] > There you go, fooling yourself with those pure math ideas again. Like > he said, _try_ it. It's amazing! It turns into a hyperbola, until you > rotate your _head_ 45 degrees too. Then it looks like a parabola again, > unless you rotate your head the wrong way, in which case it vanishes > entirely. In between, keep your head still and it turns into a square > at 1, 4, 9, 16, 25 and 36 degrees, a circle at around 31.4 degrees, and > if you keep going you get a really cool pointillist sketch of what looks > like the the Four Horsemen of the Apocalypse at 66 3/5 degrees. Applied > math rulz! >[David C. Ullrich] >> Wow - I'm surprised to see you saying these Things here. I mean >> I know you think you have protection, but you must realize >> that the NSA is listening in - you really think your guys are >> bigger than them? >LOL -- what makes you imagine my guys _aren't_ them? Ask yourself who >stands to gain the most from an unsuspected ability to crack RSA in >microseconds. That's right, the Russian mafia. Then ask yourself who the >CIA really works for. Then ask yourself who the Russian mafia really works >for. Put it all together and you get exponentially growing piles of >rabbits. >> Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, ... degr >So you knew that all along but were playing possum. Clever! What you don't >know is James's proof that distinct Fibonacci numbers are never congruent >modulo 360. You may even believe it's trivial to prove the contrary. >Here's the correct proof: suppose F_i =/= F_j but F_i = F_j modulo 360. >Then rotating a parabola F_i degrees gives the same number of rabbits as >rotating a parabola F_j degrees, and rotating a parabola F_i - F_j degrees >gives F_i - F_j rabbits. But F_i =/= F_j, so F_i - F_j is not 0. Therefore >F_i - F_j may as well be 45. But if you rotate a parabola F_i - F_j = 45 >degrees, you get a hyperbola, and a hyperbola is not 45 rabbits. QED. This is scary. I mean scary. On the one hand, the math doesn't lie. Especially when the math is connected. On the other hand, I almost feel I should point out that 89 and 39088169 appear to be... No, never mind. Have a nice day, sir. Let me know if I can be of service in any way. >> You never tire of finding _new_ ways to make a public fool >> of yourself. >Well, I tried, but I'm afraid that argument was too clear ************************ David C. Ullrich === Subject: Re: JSH: SF might work, bad luck? [David C. Ullrich, struggling to learn about parabola rotations] > ... > Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, ... degr [Tim Peters] >> So you knew that all along but were playing possum. Clever! What you >> don't know is James's proof that distinct Fibonacci numbers are never >> congruent modulo 360. You may even believe it's trivial to prove the >> contrary. Here's the correct proof: suppose F_i =/= F_j but F_i = F_j >> modulo 360. Then rotating a parabola F_i degrees gives the same number >> of rabbits as rotating a parabola F_j degrees, and rotating a parabola >> F_i - F_j degrees gives F_i - F_j rabbits. But F_i =/= F_j, so F_i - F_j >> is not 0. Therefore F_i - F_j may as well be 45. But if you rotate a >> parabola F_i - F_j = 45 degrees, you get a hyperbola, and a hyperbola is >> not 45 rabbits. QED. [David C. Ullrich] > This is scary. I mean scary. On the one hand, the math doesn't lie. > Especially when the math is connected. On the other hand, I almost > feel I should point out that 89 and 39088169 appear to be... Yes, they both _do_ appear to be congruent to 17 modulo 360 (proof: 89 = 17 + (1/5)*360; etc)-- if you're a pigeon with a hole in your brain. Do the math right and you'll see that 39088169 is actually congruent to 89 mod 360, but 89 is congruent to 88 mod 360. Finally, 88 =/= 89, because 89-88 = (sqrt(89) + sqrt(88))*(sqrt(89) - sqrt(88)), and neither factor is rational, so their product can't be 0. I was expecting you to point out that 1548008755920 and 5358359254990966640871840 are both multiples of 360. Many fall into that trap. While true, it's not really a counterexample, since you get 0 rabbits then, and a parabola doesn't have any rabbits either. The math is perfect. > No, never mind. Have a nice day, sir. Let me know if I can be of > service in any way. I suppose you could do the world a favor by letting this drop now, since I appear to be helplessly compelled to keep replying no matter _how_ foolish your arguments get === Subject: Re: JSH: SF might work, bad luck? On Sun, 11 Jun 2006 15:28:45 -0400, Tim Peters > ... >> [...] >> No, never mind. Have a nice day, sir. Let me know if I can be of >> service in any way. >I suppose you could do the world a favor by letting this drop now, since I >appear to be helplessly compelled to keep replying no matter _how_ foolish >your arguments get Yes, sir. No problem. You wish to have the last word, only too happy to comply. ************************ David C. Ullrich === Subject: Re: JSH: SF might work, bad luck? > [David C. Ullrich, struggling to learn about parabola rotations] >> ... >> Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, ... >> degr > [Tim Peters] > So you knew that all along but were playing possum. Clever! What > you > don't know is James's proof that distinct Fibonacci numbers are > never > congruent modulo 360. You may even believe it's trivial to prove > the > contrary. Here's the correct proof: suppose F_i =/= F_j but F_i > = F_j > modulo 360. Then rotating a parabola F_i degrees gives the same > number > of rabbits as rotating a parabola F_j degrees, and rotating a > parabola > F_i - F_j degrees gives F_i - F_j rabbits. But F_i =/= F_j, so > F_i - F_j > is not 0. Therefore F_i - F_j may as well be 45. But if you > rotate a > parabola F_i - F_j = 45 degrees, you get a hyperbola, and a > hyperbola is > not 45 rabbits. QED. > [David C. Ullrich] >> This is scary. I mean scary. On the one hand, the math doesn't lie. >> Especially when the math is connected. On the other hand, I almost >> feel I should point out that 89 and 39088169 appear to be... > Yes, they both _do_ appear to be congruent to 17 modulo 360 > (proof: 89 = 17 + (1/5)*360; etc)-- if you're a pigeon with a hole > in your brain. Do the math right and you'll see that 39088169 is > actually congruent to 89 mod 360, but 89 is congruent to 88 mod 360. > Finally, 88 =/= 89, because 89-88 = (sqrt(89) + > sqrt(88))*(sqrt(89) - sqrt(88)), and neither factor is rational, so > their product can't be 0. > I was expecting you to point out that 1548008755920 and > 5358359254990966640871840 are both multiples of 360. Many fall into > that trap. While true, it's not really a counterexample, since you > get 0 rabbits then, and a parabola doesn't have any rabbits either. > The math is perfect. The parabola does not have rabbits, but it does have an associated point not previously recognised: the point of intersection of its asymptotes. I call this the Harris point. It is 10 yards (360 inches!) from the latus rectum and, amazingly, slightly to one side of the parabola's axis of symmetry. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: JSH: SF might work, bad luck? <4k1g82tjdtvlu8kaurs04u201h6o1k4t68@4ax.com> <03qi82p5emir89rasd1kqvt7er4e5ca9qk@4ax.com> <4f4ktmF1fva9pU1@individual.net [David C. Ullrich, struggling to learn about parabola rotations] >> ... >> Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, ... >> degr > [Tim Peters] > So you knew that all along but were playing possum. Clever! What > you > don't know is James's proof that distinct Fibonacci numbers are > never > congruent modulo 360. You may even believe it's trivial to prove > the > contrary. Here's the correct proof: suppose F_i =/= F_j but F_i > = F_j > modulo 360. Then rotating a parabola F_i degrees gives the same > number > of rabbits as rotating a parabola F_j degrees, and rotating a > parabola > F_i - F_j degrees gives F_i - F_j rabbits. But F_i =/= F_j, so > F_i - F_j > is not 0. Therefore F_i - F_j may as well be 45. But if you > rotate a > parabola F_i - F_j = 45 degrees, you get a hyperbola, and a > hyperbola is > not 45 rabbits. QED. > [David C. Ullrich] >> This is scary. I mean scary. On the one hand, the math doesn't lie. >> Especially when the math is connected. On the other hand, I almost >> feel I should point out that 89 and 39088169 appear to be... > Yes, they both _do_ appear to be congruent to 17 modulo 360 > (proof: 89 = 17 + (1/5)*360; etc)-- if you're a pigeon with a hole > in your brain. Do the math right and you'll see that 39088169 is > actually congruent to 89 mod 360, but 89 is congruent to 88 mod 360. > Finally, 88 =/= 89, because 89-88 = (sqrt(89) + > sqrt(88))*(sqrt(89) - sqrt(88)), and neither factor is rational, so > their product can't be 0. > I was expecting you to point out that 1548008755920 and > 5358359254990966640871840 are both multiples of 360. Many fall into > that trap. While true, it's not really a counterexample, since you > get 0 rabbits then, and a parabola doesn't have any rabbits either. > The math is perfect. > The parabola does not have rabbits, but it does have an associated > point not previously recognised: the point of intersection of its > asymptotes. I call this the Harris point. It is 10 yards (360 inches!) > from the latus rectum and, amazingly, slightly to one side of the > parabola's axis of symmetry. I don't know, Clive. I'm concerned that it doesn't look professional that you measured the distance in inches. Should be radians, yes? But it's good to hear the parabola has no rabbits. I don't like them, unless of course they are made of peanut butter, because they'll make a mess digging around in my pigeon-holes. === Subject: Re: JSH: SF might work, bad luck? >> [David C. Ullrich, struggling to learn about parabola rotations] > ... > Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, > ... > degr >> [Tim Peters] >> So you knew that all along but were playing possum. Clever! >> What >> you >> don't know is James's proof that distinct Fibonacci numbers are >> never >> congruent modulo 360. You may even believe it's trivial to >> prove >> the >> contrary. Here's the correct proof: suppose F_i =/= F_j but >> F_i >> = F_j >> modulo 360. Then rotating a parabola F_i degrees gives the same >> number >> of rabbits as rotating a parabola F_j degrees, and rotating a >> parabola >> F_i - F_j degrees gives F_i - F_j rabbits. But F_i =/= F_j, so >> F_i - F_j >> is not 0. Therefore F_i - F_j may as well be 45. But if you >> rotate a >> parabola F_i - F_j = 45 degrees, you get a hyperbola, and a >> hyperbola is >> not 45 rabbits. QED. >> [David C. Ullrich] > This is scary. I mean scary. On the one hand, the math doesn't > lie. > Especially when the math is connected. On the other hand, I > almost > feel I should point out that 89 and 39088169 appear to be... >> Yes, they both _do_ appear to be congruent to 17 modulo 360 >> (proof: 89 = 17 + (1/5)*360; etc)-- if you're a pigeon with a >> hole >> in your brain. Do the math right and you'll see that 39088169 is >> actually congruent to 89 mod 360, but 89 is congruent to 88 mod >> 360. >> Finally, 88 =/= 89, because 89-88 = (sqrt(89) + >> sqrt(88))*(sqrt(89) - sqrt(88)), and neither factor is rational, >> so >> their product can't be 0. >> I was expecting you to point out that 1548008755920 and >> 5358359254990966640871840 are both multiples of 360. Many fall >> into >> that trap. While true, it's not really a counterexample, since >> you >> get 0 rabbits then, and a parabola doesn't have any rabbits >> either. >> The math is perfect. >> The parabola does not have rabbits, but it does have an associated >> point not previously recognised: the point of intersection of its >> asymptotes. I call this the Harris point. It is 10 yards (360 >> inches!) >> from the latus rectum and, amazingly, slightly to one side of the >> parabola's axis of symmetry. > I don't know, Clive. I'm concerned that it doesn't look > professional > that you measured the distance in inches. Should be radians, yes? > But it's good to hear the parabola has no rabbits. I don't like > them, > unless of course they are made of peanut butter, because they'll > make a mess digging around in my pigeon-holes. Yes, Chip, I think the distance should be measured in radians. That would be much more... erudite. So the distance would then be 2pi radians rather than 360 inches, I _think_. Anyway. Units are soooo important. If there was a convention that the parameter to the exp function was measured in units of i*pi then we would have e = e^1 = -1 Yes, e would be -1. Much simpler. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: JSH: SF might work, bad luck? <128ap1te73mhhad@corp.supernews.com> <4k1g82tjdtvlu8kaurs04u201h6o1k4t68@4ax.com> <03qi82p5emir89rasd1kqvt7er4e5ca9qk@4ax.com> > [...] >> Try it. Rotate a parabola and see what you get. >> If it seems too abstract, post what you get. >> Like, rotate a parabola 45 degrees. >> Post the result. > [David C. Ullrich] > Ok. If you rotate a parbola 45 degrees the result > is a parabola. > [Tim Peters] >> There you go, fooling yourself with those pure math ideas again. Like >> he said, _try_ it. It's amazing! It turns into a hyperbola, until you >> rotate your _head_ 45 degrees too. Then it looks like a parabola again, >> unless you rotate your head the wrong way, in which case it vanishes >> entirely. In between, keep your head still and it turns into a square >> at 1, 4, 9, 16, 25 and 36 degrees, a circle at around 31.4 degrees, and >> if you keep going you get a really cool pointillist sketch of what looks >> like the the Four Horsemen of the Apocalypse at 66 3/5 degrees. Applied >> math rulz! > [David C. Ullrich] > Wow - I'm surprised to see you saying these Things here. I mean > I know you think you have protection, but you must realize > that the NSA is listening in - you really think your guys are > bigger than them? > LOL -- what makes you imagine my guys _aren't_ them? Ask yourself who > stands to gain the most from an unsuspected ability to crack RSA in > microseconds. That's right, the Russian mafia. Then ask yourself who the > CIA really works for. Then ask yourself who the Russian mafia really works > for. Put it all together and you get exponentially growing piles of > rabbits. > Next you'll be telling us what happens at 1, 1, 2, 3, 5, 8, ... degr > So you knew that all along but were playing possum. Clever! What you don't > know is James's proof that distinct Fibonacci numbers are never congruent > modulo 360. You may even believe it's trivial to prove the contrary. > Here's the correct proof: suppose F_i =/= F_j but F_i = F_j modulo 360. > Then rotating a parabola F_i degrees gives the same number of rabbits as > rotating a parabola F_j degrees, and rotating a parabola F_i - F_j degrees > gives F_i - F_j rabbits. But F_i =/= F_j, so F_i - F_j is not 0. Therefore > F_i - F_j may as well be 45. But if you rotate a parabola F_i - F_j = 45 > degrees, you get a hyperbola, and a hyperbola is not 45 rabbits. QED. > You never tire of finding _new_ ways to make a public fool > of yourself. > Well, I tried, but I'm afraid that argument was too clear David's mocking of JSH, er... okay I didn't quite follow that, but: > So you knew that all along but were playing possum. Clever! > What you don't know is James's proof that distinct Fibonacci > numbers are never congruent modulo 360. Oh yeah? Well, at this very moment I've set my super-duper computer to finding the very last 361 Fibonacci numbers, and when I do I will force them to fly as one flock into a number of carefully prepared pigeon-holes. The pigeon residues will be excrementable. Oh, the avianity! Blame yourselves now and avoid the rush. -- c === Subject: It looks simple but.. I have a nonlinear differential equation to solve. I need a symbolic equation for x(t) that fits this equation x'(t)+ a*x(t)^2 = b x(0)=0 // initial position x'(0)=b // iniitia velocity Am I missing something or does this need be solved numerically? I see examples in the text books that look similar such as y'=y^2, but no integrating factors are obvious for my equation. . This is not homework. x is the position of an object and x' is the velocity of an object that compresses a spring. The gain of the system drops a function of the increase in force All of these parameter are folded into the constants a and b. Peter Nachtwey === Subject: Re: It looks simple but.. >I have a nonlinear differential equation to solve. I need a symbolic > equation for x(t) that fits this equation > x'(t)+ a*x(t)^2 = b > x(0)=0 // initial position > x'(0)=b // iniitia velocity > Am I missing something or does this need be solved numerically? I see > examples in the text books that look similar such as y'=y^2, but no > integrating factors are obvious for my equation. . > This is not homework. x is the position of an object and x' is the > velocity of an object that compresses a spring. The gain of the system > drops a function of the increase in force > All of these parameter are folded into the constants a and b. Write it as dx/dt = b - a x^2 and rewrite as 1 / ( b - a x^2 ) dx = dt . Then integrate both sides: atanh( x sqrt(a/b) ) / sqrt(a b) - 0 = t - 0 providing a and b have the same sign. Inversion gives x(t) = sqrt(b/a) tanh( t sqrt(a b) ) check: x'(t) = sqrt(b/a) sqrt(a b) [ 1 - tanh^2(t sqrt(a b)) ] = b [ 1 - tanh^2(t sqrt(a b)) ] so x'(0) = b Dirk Vdm === Subject: Re: It looks simple but.. <_qGig.475387$7V4.12390452@phobos.telenet-ops.beI have a nonlinear differential equation to solve. I need a symbolic > equation for x(t) that fits this equation > x'(t)+ a*x(t)^2 = b > x(0)=0 // initial position > x'(0)=b // iniitia velocity > Am I missing something or does this need be solved numerically? I see > examples in the text books that look similar such as y'=y^2, but no > integrating factors are obvious for my equation. . > This is not homework. x is the position of an object and x' is the > velocity of an object that compresses a spring. The gain of the system > drops a function of the increase in force > All of these parameter are folded into the constants a and b. > Write it as > dx/dt = b - a x^2 > and rewrite as > 1 / ( b - a x^2 ) dx = dt . > Then integrate both sides: > atanh( x sqrt(a/b) ) / sqrt(a b) - 0 = t - 0 > providing a and b have the same sign. > Inversion gives > x(t) = sqrt(b/a) tanh( t sqrt(a b) ) > check: > x'(t) = sqrt(b/a) sqrt(a b) [ 1 - tanh^2(t sqrt(a b)) ] > = b [ 1 - tanh^2(t sqrt(a b)) ] > so > x'(0) = b > Dirk Vdm didn't solve that. I think I should look into Mathematica. I will now compare the two solutions. They look similar. Peter Nachtwey === Subject: Re: It looks simple but.. >>I have a nonlinear differential equation to solve. I need a symbolic >> equation for x(t) that fits this equation >> x'(t)+ a*x(t)^2 = b >> x(0)=0 // initial position >> x'(0)=b // iniitia velocity >> Am I missing something or does this need be solved numerically? I see >> examples in the text books that look similar such as y'=y^2, but no >> integrating factors are obvious for my equation. . >> This is not homework. x is the position of an object and x' is the >> velocity of an object that compresses a spring. The gain of the system >> drops a function of the increase in force >> All of these parameter are folded into the constants a and b. >> Write it as >> dx/dt = b - a x^2 >> and rewrite as >> 1 / ( b - a x^2 ) dx = dt . >> Then integrate both sides: >> atanh( x sqrt(a/b) ) / sqrt(a b) - 0 = t - 0 >> providing a and b have the same sign. >> Inversion gives >> x(t) = sqrt(b/a) tanh( t sqrt(a b) ) >> check: >> x'(t) = sqrt(b/a) sqrt(a b) [ 1 - tanh^2(t sqrt(a b)) ] >> = b [ 1 - tanh^2(t sqrt(a b)) ] >> so >> x'(0) = b >> Dirk Vdm > didn't solve that. I think I should look into Mathematica. > I will now compare the two solutions. They look similar. Note that the solution is valid if a and b have the same sign. If they have opposite signs, then the integration obviously gives atan( x sqrt(-a/b) ) / sqrt(-a b) - 0 = t - 0 Inversion gives then x(t) = - sqrt(-b/a) tan( t sqrt(-a b) ) You can check that this also gives x'(0) = b Dirk Vdm === Subject: Re: It looks simple but.. <_qGig.475387$7V4.12390452@phobos.telenet-ops.be >I have a nonlinear differential equation to solve. I need a symbolic > equation for x(t) that fits this equation > > x'(t)+ a*x(t)^2 = b > x(0)=0 // initial position > x'(0)=b // iniitia velocity > > Am I missing something or does this need be solved numerically? I see > examples in the text books that look similar such as y'=y^2, but no > integrating factors are obvious for my equation. . > > This is not homework. x is the position of an object and x' is the > velocity of an object that compresses a spring. The gain of the system > drops a function of the increase in force > All of these parameter are folded into the constants a and b. > Write it as > dx/dt = b - a x^2 > and rewrite as > 1 / ( b - a x^2 ) dx = dt . > Then integrate both sides: > atanh( x sqrt(a/b) ) / sqrt(a b) - 0 = t - 0 > providing a and b have the same sign. > Inversion gives > x(t) = sqrt(b/a) tanh( t sqrt(a b) ) > check: > x'(t) = sqrt(b/a) sqrt(a b) [ 1 - tanh^2(t sqrt(a b)) ] > = b [ 1 - tanh^2(t sqrt(a b)) ] > so > x'(0) = b > Dirk Vdm > didn't solve that. I think I should look into Mathematica. Maple gets the same solution as Mathematica. (The issue of Maple vs. Mathematica is almost a religious war.) R.G. Vickson > I will now compare the two solutions. They look similar. > Peter Nachtwey === Subject: Re: It looks simple but.. | >I have a nonlinear differential equation to solve. I need a symbolic | > equation for x(t) that fits this equation | > | > x'(t)+ a*x(t)^2 = b | > x(0)=0 // initial position | > x'(0)=b // iniitia velocity | > | > Am I missing something or does this need be solved numerically? I see | > examples in the text books that look similar such as y'=y^2, but no | > integrating factors are obvious for my equation. . | > | > This is not homework. x is the position of an object and x' is the | > velocity of an object that compresses a spring. The gain of the system | > drops a function of the increase in force | > All of these parameter are folded into the constants a and b. | > | > Write it as | > dx/dt = b - a x^2 | > and rewrite as | > 1 / ( b - a x^2 ) dx = dt . | > | > Then integrate both sides: | > atanh( x sqrt(a/b) ) / sqrt(a b) - 0 = t - 0 | > providing a and b have the same sign. | > Inversion gives | > x(t) = sqrt(b/a) tanh( t sqrt(a b) ) | > | > check: | > x'(t) = sqrt(b/a) sqrt(a b) [ 1 - tanh^2(t sqrt(a b)) ] | > = b [ 1 - tanh^2(t sqrt(a b)) ] | > so | > x'(0) = b | > | > Dirk Vdm | > didn't solve that. I think I should look into Mathematica. | | Maple gets the same solution as Mathematica. (The issue of Maple vs. | Mathematica is almost a religious war.) | | R.G. Vickson It looks like I am going to have to take sides. My Mathcad isn't hacking it and I am not happy with it. . Even with all the help, I can't get Mathcad to duplicate the Maple's or Mathematica's solutions. :( I do a lot of symbolic processing. I need the best package at symbolic processing. I have wasted much of a day on this problem. After wasting a couple days I will have justified the cost of one of the better symbolic processing packages. This problem has convinced me I need to 'find new cheese' ( see 'who moved my cheese' ) . I am starting my research on the best symbolic math package. I have learned a couple of things with this thread. Peter Nachtwey === Subject: Re: It looks simple but.. <_qGig.475387$7V4.12390452@phobos.telenet-ops.be >I have a nonlinear differential equation to solve. I need a symbolic > equation for x(t) that fits this equation > > x'(t)+ a*x(t)^2 = b > x(0)=0 // initial position > x'(0)=b // iniitia velocity > > Am I missing something or does this need be solved numerically? I see > examples in the text books that look similar such as y'=y^2, but no > integrating factors are obvious for my equation. . > > This is not homework. x is the position of an object and x' is the > velocity of an object that compresses a spring. The gain of the system > drops a function of the increase in force > All of these parameter are folded into the constants a and b. > > Write it as > dx/dt = b - a x^2 > and rewrite as > 1 / ( b - a x^2 ) dx = dt . > > Then integrate both sides: > atanh( x sqrt(a/b) ) / sqrt(a b) - 0 = t - 0 > providing a and b have the same sign. > Inversion gives > x(t) = sqrt(b/a) tanh( t sqrt(a b) ) > > check: > x'(t) = sqrt(b/a) sqrt(a b) [ 1 - tanh^2(t sqrt(a b)) ] > = b [ 1 - tanh^2(t sqrt(a b)) ] > so > x'(0) = b > > Dirk Vdm > didn't solve that. I think I should look into Mathematica. > Maple gets the same solution as Mathematica. (The issue of Maple vs. > Mathematica is almost a religious war.) No it isn't. It's more important than that. 8-) --- Christopher Heckman, who is about to have a price put on his head by Muslims everywhere ... === Subject: Re: It looks simple but.. > I have a nonlinear differential equation to solve. I need a symbolic > equation for x(t) that fits this equation > x'(t)+ a*x(t)^2 = b > x(0)=0 // initial position > x'(0)=b // iniitia velocity mathematica finds that x(t) is a hyperbolic tangent, viz. x(t) = sqrt(b) / sqrt(a) tanh( sqrt(a) sqrt(b) t) vale, rip -- NB eddress is r i p 1 AT c o m c a s t DOT n e t === Subject: Re: OT: Soduko takes the country by storm! unsolvable ones published with a reward to solve them. Who did that? IIRC the math community is waiting eagerly for someone to come up with a natural example of an unsolvable problem. It would give the logicians a little breathing room (they're a dying breed in U.S. math departments). Anand Pillay left UIUC, signaling the beginning of the end for that formerly venerable logic community. TG === Subject: Re: OT: Soduko takes the country by storm! > Who did that? You jumped a bit past the clue so here's another. 15 puzzle. === Subject: Re: OT: Soduko takes the country by storm! >> Who did that? > You jumped a bit past the clue so here's another. > 15 puzzle. Mathematicians consider an impossibility proof to be a solution to the problem. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: OT: Soduko takes the country by storm! <8YGig.4223$lf4.2228@newsread1.news.pas.earthlink.net> problem. Dave, what do you mean by that? === Subject: Re: OT: Soduko takes the country by storm! >> Mathematicians consider an impossibility proof to be a solution to the >> problem. > Dave, what do you mean by that? In mathematics, a question of the form Find an X that satisfies Y is always understood to mean Find an X that satisfies Y, or prove that no such X exists. When the question is properly understood, an impossibility proof is obviously a solution. Why would anyone keep looking for an X when it is known that no such object exists? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: OT: Soduko takes the country by storm! <8YGig.4223$lf4.2228@newsread1.news.pas.earthlink.net> always understood to mean Find an X that satisfies Y, or prove that no > such X exists. Uhhh, of course. I meant to say that thus far, nobody has produced a natural example of an undecidable math statement (in the Godel sense). Therefore, the incompleteness theorem is not really of general intellectual interest (g.i.i.). tg === Subject: Re: OT: Soduko takes the country by storm! >Uhhh, of course. I meant to say that thus far, nobody has produced a >natural example of an undecidable math statement (in the Godel sense). Undecidable in what formal system? >Therefore, the incompleteness theorem is not really of general >intellectual interest (g.i.i.). The Paris-Harrington principle: For any positive integers n, k, m we can find l with the following property: if we color each of the n element subsets of {1, 2, 3,..., l} with one of k colors, then we can find a subset Y with at least m elements, such that all n element subsets of Y have the same color, and the number of elements of Y is at least the smallest element of Y. This is unprovable in first-order Peano arithmetic. See . Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: OT: Soduko takes the country by storm! >Uhhh, of course. I meant to say that thus far, nobody has produced a >natural example of an undecidable math statement (in the Godel sense). > Undecidable in what formal system? >Therefore, the incompleteness theorem is not really of general >intellectual interest (g.i.i.). > The Paris-Harrington principle: > For any positive integers n, k, m we can find l with the following > property: if we color each of the n element subsets of {1, 2, 3,..., l} > with one of k colors, then we can find a subset Y with at least m > elements, such that all n element subsets of Y have the same color, and > the number of elements of Y is at least the smallest element of Y. > This is unprovable in first-order Peano arithmetic. See > . No matter what happens on and to Usenet, I'm glad that there are people like you, Robert, who can scatter absolute nuggets such as the above into the stream. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: OT: Soduko takes the country by storm! natural example of an undecidable math statement (in the Godel sense). > Undecidable in what formal system? Obviously I meant ZFC, so your example is worthless. === Subject: Re: OT: Soduko takes the country by storm! [ removing rec.music.classical.guitar from the Newsgroups line] >Uhhh, of course. I meant to say that thus far, nobody has produced a >natural example of an undecidable math statement (in the Godel sense). > Undecidable in what formal system? > Obviously I meant ZFC, so your example is worthless. The Continuum Hypothesis, then. Or the Word Problem in groups. There is a finitely presented group G such that there is no algorithm to decide whether two words in the generators of G represent the same element of G. It follows that there exist two words in the generators of G such that the question of whether they represent the same element of G is undecidable in ZFC. Similarly for other algorithmically undecidable problems. Robert Israel israel@math.MyUniversity'sInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: OT: Soduko takes the country by storm! > Tommy Grand -snip- > This is unprovable in first-order Peano arithmetic. See > . > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Hi Robert, (and the rest of sci.math and rec.music.classical.guitar) Please note this is a cross posted troll by one Tommy Grand. He is resident of rec.music.classical.guitar who is posting to both groups as part of a game he's playing. Neither of us have posted in sci.math before and have no reason to do so as you can see. Sorry these flame games spilled over into sci.math. === Subject: Re: OT: Soduko takes the country by storm! > Mathematicians consider an impossibility proof to be a solution to the > problem. > Dave, what do you mean by that? He means that when you prove that a puzzle is imposible to solve, you have in fact found the set of all possible solutions, the empty set, and proved that there are no other solutions. That's far more significant than just finding one solution (unless you can also accompany that with a proof that it is unique). Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: OT: Soduko takes the country by storm! > Dave, what do you mean by that? Tic Tac D'oh! Hey, you want a piece of me? Takes 2 2 Tango... === Subject: Surrogate factoring, dead idea Last postings on this subject I was in dumb denial about the failure of my latest ideas with what I call surrogate factoring as I accused two posters of lying about the calculation. I was wrong, they were right. After several years of playing with this idea of looking for one factorization to pull out another, I've exhausted all the simple approaches--as I like to use basic algebra--that I can think of, and, yet again, as I've said this before, and then went after it again, I think that my surrogate factoring is is a dead idea. The algebra always loops back to the number to be factored, with the factorization dependent on it, or you get very low odds of factoring. It might make sense to most of you to just give up after years, and at the moment it makes sense to me. But it is a hobby, and I do play with very simple algebra, where I can still screw up, so it's possible down the line, I may be at it again. And, oh yeah, consider that I use VERY SIMPLE ALGEBRA and routinely screw up, usually because I want something to be true when it isn't. This factoring is practical in that people would be very interested if I succeeded, so the reality that I haven't comes up quickly because people do care about a practical answer. In contrast, in pure math areas, some mathematician can get deluded with VERY COMPLEX IDEAS and want something badly, it goes to some reviewer who just needs to get through the tedium of a review, and if no one cares, why should anyone know of a major screw-up? Truth is that when one person wants something badly, that person can convince himself or herself of things that are completely wrong. And unless they have a strong reason to check--like in applied mathematical areas--people who look over such work can miss even glaring errors because it just is not that important to them, and they may even trust the person giving the information as brilliant. I emphasize, people make mistakes. Look at how I can go on with simple algebra mistakes. Computer checking is the only way to remove the human element. It's the only way to protect from mathematicians who desperately want a solution, being checked by other mathematicians who want them to be right, in areas where there isn't a lot of money at stake like factoring, or the real world to tell what is correct, like in other applied mathematics. Sure mathematicians LIKE the current system. Why shouldn't they? If you knew your colleague aka buddy was going to be looking over your research, and in many ways had the final word, and as a result you could get thousands of dollars of research money, what's not to like? Versus some heartless computer that will slash your proofs to dust. What's not to like about the current system for today's pure mathematician? James Harris === Subject: Re: Surrogate factoring, dead idea > And unless they have a strong reason to check--like in applied > mathematical areas--people who look over such work can miss even > glaring errors because it just is not that important to them > James Harris === Subject: Re: Surrogate factoring, dead idea >> And unless they have a strong reason to check--like in applied >> mathematical areas--people who look over such work can miss even >> glaring errors because it just is not that important to them Or they all just wanted a good laugh === Subject: Re: Surrogate factoring, dead idea > I was wrong, they were right. Have the world's financial institutions been informed? Trading pretty muched ceased last week as we waited for your announcement on this. At last, the world can move on. > Last postings on this subject I was in dumb denial about the failure of > my latest ideas with what I call surrogate factoring as I accused two > posters of lying about the calculation. > After several years of playing with this idea of looking for one > factorization to pull out another, I've exhausted all the simple > approaches--as I like to use basic algebra--that I can think of, and, > yet again, as I've said this before, and then went after it again, I > think that my surrogate factoring is is a dead idea. > The algebra always loops back to the number to be factored, with the > factorization dependent on it, or you get very low odds of factoring. > It might make sense to most of you to just give up after years, and at > the moment it makes sense to me. But it is a hobby, and I do play with > very simple algebra, where I can still screw up, so it's possible down > the line, I may be at it again. > And, oh yeah, consider that I use VERY SIMPLE ALGEBRA and routinely > screw up, usually because I want something to be true when it isn't. > This factoring is practical in that people would be very interested if > I succeeded, so the reality that I haven't comes up quickly because > people do care about a practical answer. > In contrast, in pure math areas, some mathematician can get deluded > with VERY COMPLEX IDEAS and want something badly, it goes to some > reviewer who just needs to get through the tedium of a review, and if > no one cares, why should anyone know of a major screw-up? > Truth is that when one person wants something badly, that person can > convince himself or herself of things that are completely wrong. > And unless they have a strong reason to check--like in applied > mathematical areas--people who look over such work can miss even > glaring errors because it just is not that important to them, and they > may even trust the person giving the information as brilliant. > I emphasize, people make mistakes. Look at how I can go on with simple > algebra mistakes. > Computer checking is the only way to remove the human element. > It's the only way to protect from mathematicians who desperately want a > solution, being checked by other mathematicians who want them to be > right, in areas where there isn't a lot of money at stake like > factoring, or the real world to tell what is correct, like in other > applied mathematics. > Sure mathematicians LIKE the current system. Why shouldn't they? > If you knew your colleague aka buddy was going to be looking over > your research, and in many ways had the final word, and as a result you > could get thousands of dollars of research money, what's not to like? > Versus some heartless computer that will slash your proofs to dust. > What's not to like about the current system for today's pure > mathematician? > James Harris === Subject: Re: Surrogate factoring, dead idea > I was wrong, they were right. Have the world's financial institutions been informed? Trading pretty muched ceased last week as we waited for your announcement on this. At last, the world can move on. > Last postings on this subject I was in dumb denial about the failure of > my latest ideas with what I call surrogate factoring as I accused two > posters of lying about the calculation. > After several years of playing with this idea of looking for one > factorization to pull out another, I've exhausted all the simple > approaches--as I like to use basic algebra--that I can think of, and, > yet again, as I've said this before, and then went after it again, I > think that my surrogate factoring is is a dead idea. > The algebra always loops back to the number to be factored, with the > factorization dependent on it, or you get very low odds of factoring. > It might make sense to most of you to just give up after years, and at > the moment it makes sense to me. But it is a hobby, and I do play with > very simple algebra, where I can still screw up, so it's possible down > the line, I may be at it again. > And, oh yeah, consider that I use VERY SIMPLE ALGEBRA and routinely > screw up, usually because I want something to be true when it isn't. > This factoring is practical in that people would be very interested if > I succeeded, so the reality that I haven't comes up quickly because > people do care about a practical answer. > In contrast, in pure math areas, some mathematician can get deluded > with VERY COMPLEX IDEAS and want something badly, it goes to some > reviewer who just needs to get through the tedium of a review, and if > no one cares, why should anyone know of a major screw-up? > Truth is that when one person wants something badly, that person can > convince himself or herself of things that are completely wrong. > And unless they have a strong reason to check--like in applied > mathematical areas--people who look over such work can miss even > glaring errors because it just is not that important to them, and they > may even trust the person giving the information as brilliant. > I emphasize, people make mistakes. Look at how I can go on with simple > algebra mistakes. > Computer checking is the only way to remove the human element. > It's the only way to protect from mathematicians who desperately want a > solution, being checked by other mathematicians who want them to be > right, in areas where there isn't a lot of money at stake like > factoring, or the real world to tell what is correct, like in other > applied mathematics. > Sure mathematicians LIKE the current system. Why shouldn't they? > If you knew your colleague aka buddy was going to be looking over > your research, and in many ways had the final word, and as a result you > could get thousands of dollars of research money, what's not to like? > Versus some heartless computer that will slash your proofs to dust. > What's not to like about the current system for today's pure > mathematician? > James Harris === Subject: Re: Surrogate factoring, dead idea Discussion, linux) > This factoring is practical in that people would be very interested if > I succeeded, so the reality that I haven't comes up quickly because > people do care about a practical answer. > In contrast, in pure math areas, some mathematician can get deluded > with VERY COMPLEX IDEAS and want something badly, it goes to some > reviewer who just needs to get through the tedium of a review, and if > no one cares, why should anyone know of a major screw-up? That's a pretty compelling analysis of the difference between factoring and most pure mathematics. But let's play what if. What if the real difference is that you can check your factoring algorithms and see that they don't work? You don't know how to check your other results so you believe that they're right and everyone else is wrong (and has been wrong for over a century). Is that possibly what's going on here? -- Quincy, would you rather do epistemology or conceptual analysis? You know what? I'd rather fight on an aircraft carrier.... And Mama and Baba (Papa) would fight on an aircraft carrier, too. -- Quincy P. Hughes, age 3 1/2 === Subject: Re: JSH: Surrogate factoring, dead idea [added JSH: to subject, spared sci.crypt and alt.math] [jstevh@msn.com] > Last postings on this subject I was in dumb denial about the failure of > my latest ideas with what I call surrogate factoring as I accused two > posters of lying about the calculation. > I was wrong, they were right. Fat lot of comfort that's going to be to the survivors if some loon acts on your statements that Decker & I deserved to be killed for lying here. You went way over the line there. Dig yourself a deeper hole in response. In any case, I was impressed that you were able to shut up after that round of crazy accusations long enough to figure out the problem for yourself. That's better than you usually manage at the end of a cycle (you usually escalate public ranting instead), and I hope that improvement lasts. > After several years of playing with this idea of looking for one > factorization to pull out another, I've exhausted all the simple > approaches--as I like to use basic algebra--that I can think of, and, > yet again, as I've said this before, and then went after it again, I > think that my surrogate factoring is is a dead idea. > The algebra always loops back to the number to be factored, with the > factorization dependent on it, or you get very low odds of factoring. > It might make sense to most of you to just give up after years, and at > the moment it makes sense to me. But it is a hobby, and I do play with > very simple algebra, where I can still screw up, so it's possible down > the line, I may be at it again. If you were able to talk about it without going nuts, why not? Nobody objects to exploring ideas, it's the endless bogus claims of triumph against all reason that bother people who prefer some, well, _proof_ with their proofs. > And, oh yeah, consider that I use VERY SIMPLE ALGEBRA and routinely > screw up, usually because I want something to be true when it isn't. > This factoring is practical in that people would be very interested if > I succeeded, so the reality that I haven't comes up quickly because > people do care about a practical answer. > In contrast, in pure math areas, some mathematician can get deluded > with VERY COMPLEX IDEAS and want something badly, it goes to some > reviewer who just needs to get through the tedium of a review, and if > no one cares, why should anyone know of a major screw-up? > Truth is that when one person wants something badly, that person can > convince himself or herself of things that are completely wrong. That's actually rare in mathematics. > And unless they have a strong reason to check--like in applied > mathematical areas--people who look over such work can miss even > glaring errors because it just is not that important to them, and they > may even trust the person giving the information as brilliant. > I emphasize, people make mistakes. Look at how I can go on with simple > algebra mistakes. Which are pointed out to you almost instantly when you post them. If you read threads other than your own, you'd see how it normally goes: person A moves on. You're one of the few who fights like a maniac to hold on to his mistakes. The few who do that routinely are conventionally called cranks. > Computer checking is the only way to remove the human element. Doesn't actually help: Decker and I _both_ used computer derivation to show you what was wrong this time, and your response was just liar, liar, liar. _That's_ the human element right there: a person will hold on to a crazy belief for just as long as they prefer it to reality, and technology can't change that. > It's the only way to protect from mathematicians who desperately want a > solution, being checked by other mathematicians who want them to be > right, in areas where there isn't a lot of money at stake like > factoring, or the real world to tell what is correct, like in other > applied mathematics. Except there's no reason to believe, or even suspect, that this is a significant problem in mathematical reality. (Hint: you haven't actually found a problem in ideal theory -- you're confused about that.) > Sure mathematicians LIKE the current system. Why shouldn't they? > If you knew your colleague aka buddy was going to be looking over > your research, and in many ways had the final word, and as a result you > could get thousands of dollars of research money, what's not to like? > Versus some heartless computer that will slash your proofs to dust. > What's not to like about the current system for today's pure > mathematician? If you get a job doing pure-math research, you can tell us -- if you do, you're going to find a _lot_ of things you hate about it. === Subject: Re: Surrogate factoring, dead idea > Last postings on this subject I was in dumb denial about the failure of > my latest ideas with what I call surrogate factoring as I accused two > posters of lying about the calculation. > I was wrong, they were right. I don't see any apolgies for calling people liars, threatening to get them fired from their jobs, or calling for their blood. > After several years of playing with this idea of looking for one > factorization to pull out another, I've exhausted all the simple > approaches--as I like to use basic algebra--that I can think of, and, > yet again, as I've said this before, and then went after it again, I > think that my surrogate factoring is is a dead idea. > The algebra always loops back to the number to be factored, with the > factorization dependent on it, or you get very low odds of factoring. > It might make sense to most of you to just give up after years, Or even weeks. If you're a pro, you can see most of the dead ends of research without having to actually encounter them. > and at > the moment it makes sense to me. But it is a hobby, and I do play with > very simple algebra, where I can still screw up, so it's possible down > the line, I may be at it again. The odds are in favor of it. They're currently 900:1 in favor of it at Vegas. > And, oh yeah, consider that I use VERY SIMPLE ALGEBRA and routinely > screw up, usually because I want something to be true when it isn't. > This factoring is practical in that people would be very interested if > I succeeded, so the reality that I haven't comes up quickly because > people do care about a practical answer. > In contrast, in pure math areas, some mathematician can get deluded > with VERY COMPLEX IDEAS and want something badly, it goes to some > reviewer who just needs to get through the tedium of a review, and if > no one cares, why should anyone know of a major screw-up? Mathematics is the process of examining very complex ideas and seeing what follows as a result. Having not seen any (say, topology, abstract algebra, analysis), you don't know this. > Truth is that when one person wants something badly, that person can > convince himself or herself of things that are completely wrong. > And unless they have a strong reason to check--like in applied > mathematical areas--people who look over such work can miss even > glaring errors because it just is not that important to them, and they > may even trust the person giving the information as brilliant. > I emphasize, people make mistakes. Look at how I can go on with simple > algebra mistakes. > Computer checking is the only way to remove the human element. > It's the only way to protect from mathematicians who desperately want a > solution, being checked by other mathematicians who want them to be > right, in areas where there isn't a lot of money at stake like > factoring, or the real world to tell what is correct, like in other > applied mathematics. > Sure mathematicians LIKE the current system. Why shouldn't they? You haven't talked to many mathematicians lately. If a professional mathematician stumbles on a proof that seems to easy, he/she doesn't immediately announce it to the world; he/she will think about it for several weeks and look for any flaws in the argument/proof. Since the hardest mistakes to find are usually one's own, he/she will ask several colleges to look at a proof. Finally, when a paper is written up, it is checked --- even more thoroughly, because the reviewer usually won't have seen the result ahead of time. (And in fact, the ethical thing to do if the reviewer _has_ seen the result is to reject it, which is always a possibility.) > If you knew your colleague aka buddy was going to be looking over > your research, and in many ways had the final word, and as a result you > could get thousands of dollars of research money, what's not to like? But you can't guarantee that. The reviewer may be your worst enemy. For instance, who looked over your Advanced Polynomial Factoring paper that you submitted to that journal in Oklahoma? Answer: You don't know. Only the editor and the reviewer (if there was one) should know. > Versus some heartless computer that will slash your proofs to dust. When I was given a paper to review for the Journal of Graph Theory, I did slash [it] to dust. Well, that was the attitude, but since the results were correct, and the proofs were actual mathematical proofs (after making a few minor, obvious changes), the paper survived. And this was despite the fact that the author can be a real jerk at times (on days ending with Y). > What's not to like about the current system for today's pure > mathematician? How much pure mathematics do you know? Close to zero. Argument from ignorance again. --- Christopher Heckman === Subject: Re: Surrogate factoring, dead idea >> Last postings on this subject I was in dumb denial about the failure of >> my latest ideas with what I call surrogate factoring as I accused two >> posters of lying about the calculation. >> I was wrong, they were right. > I don't see any apolgies for calling people liars, threatening to get > them fired from their jobs, or calling for their blood. >> After several years of playing with this idea of looking for one >> factorization to pull out another, I've exhausted all the simple >> approaches--as I like to use basic algebra--that I can think of, and, >> yet again, as I've said this before, and then went after it again, I >> think that my surrogate factoring is is a dead idea. >> The algebra always loops back to the number to be factored, with the >> factorization dependent on it, or you get very low odds of factoring. >> It might make sense to most of you to just give up after years, > Or even weeks. If you're a pro, you can see most of the dead ends of > research without having to actually encounter them. >> and at >> the moment it makes sense to me. But it is a hobby, and I do play with >> very simple algebra, where I can still screw up, so it's possible down >> the line, I may be at it again. > The odds are in favor of it. They're currently 900:1 in favor of it at > Vegas. >> And, oh yeah, consider that I use VERY SIMPLE ALGEBRA and routinely >> screw up, usually because I want something to be true when it isn't. >> This factoring is practical in that people would be very interested if >> I succeeded, so the reality that I haven't comes up quickly because >> people do care about a practical answer. >> In contrast, in pure math areas, some mathematician can get deluded >> with VERY COMPLEX IDEAS and want something badly, it goes to some >> reviewer who just needs to get through the tedium of a review, and if >> no one cares, why should anyone know of a major screw-up? > Mathematics is the process of examining very complex ideas and seeing > what follows as a result. Having not seen any (say, topology, abstract > algebra, analysis), you don't know this. >> Truth is that when one person wants something badly, that person can >> convince himself or herself of things that are completely wrong. >> And unless they have a strong reason to check--like in applied >> mathematical areas--people who look over such work can miss even >> glaring errors because it just is not that important to them, and they >> may even trust the person giving the information as brilliant. >> I emphasize, people make mistakes. Look at how I can go on with simple >> algebra mistakes. >> Computer checking is the only way to remove the human element. >> It's the only way to protect from mathematicians who desperately want a >> solution, being checked by other mathematicians who want them to be >> right, in areas where there isn't a lot of money at stake like >> factoring, or the real world to tell what is correct, like in other >> applied mathematics. >> Sure mathematicians LIKE the current system. Why shouldn't they? > You haven't talked to many mathematicians lately. If a professional > mathematician stumbles on a proof that seems to easy, he/she doesn't > immediately announce it to the world; he/she will think about it for > several weeks and look for any flaws in the argument/proof. Since the > hardest mistakes to find are usually one's own, he/she will ask several > colleges to look at a proof. Finally, when a paper is written up, it is > checked --- even more thoroughly, because the reviewer usually won't > have seen the result ahead of time. (And in fact, the ethical thing to > do if the reviewer _has_ seen the result is to reject it, which is > always a possibility.) >> If you knew your colleague aka buddy was going to be looking over >> your research, and in many ways had the final word, and as a result you >> could get thousands of dollars of research money, what's not to like? > But you can't guarantee that. The reviewer may be your worst enemy. > For instance, who looked over your Advanced Polynomial Factoring > paper that you submitted to that journal in Oklahoma? Answer: You don't > know. Only the editor and the reviewer (if there was one) should know. >> Versus some heartless computer that will slash your proofs to dust. > When I was given a paper to review for the Journal of Graph Theory, I > did slash [it] to dust. Well, that was the attitude, but since the > results were correct, and the proofs were actual mathematical proofs > (after making a few minor, obvious changes), the paper survived. And > this was despite the fact that the author can be a real jerk at times > (on days ending with Y). >> What's not to like about the current system for today's pure >> mathematician? > How much pure mathematics do you know? Close to zero. > Argument from ignorance again. JSH is a troll in algebra challanged clothing. God does not play dice with the universe, and he is a mathematition, and he will get the troll JSH === Subject: Re: Surrogate factoring, dead idea > Computer checking is the only way to remove the human element. Did you already figure out how to explain to a computer what 7 factors through means? Or, at least, how to explain it to us? Jose Carlos Santos === Subject: Re: Surrogate factoring, dead idea > Computer checking is the only way to remove the human element. Yeah, right. Tell me, who programs the computers? === Subject: Re: Surrogate factoring, dead idea |> Computer checking is the only way to remove the human element. | | Yeah, right. Tell me, who programs the computers? | Non-Humans, you IDIOT!! Bwahahahahaha -- Ed. ----------------------------------------------------- hex->bin->b64 79DC1E8ABD40B42A2671AB2D0E8B4D7A === Subject: Re: Surrogate factoring, dead idea > Computer checking is the only way to remove the human element. > Yeah, right. Tell me, who programs the computers? Weak. Math people toss that up all the time like because computer programmers can make mistakes in the programs computer checking of mathematical arguments claimed to be proofs can't work. Trouble is, even AFTER a computer checks, a human being can check as well. And the computer will make less errors as time goes on and the programs are steadily improved upon, until it is perfect. As an example, consider compilers. They are in many ways checkers, and can catch a lot of programming errors, though not all, and over time they've gotten better and better. A mathematical proof is kind of like a computer program in that it goes step by step, but perfectly, with absolutely no errors. A proof is a claim of having a mathematical proof--a claim of absolute perfection. Computers CAN be perfect, but human beings just keep screwing things up. It's a safer bet on computers checking mathematical arguments plus human beings, if needed, versus just letting some dude eye-ball some arguments and claim that they are correct. James Harris === Subject: Re: Surrogate factoring, dead idea > Computer checking is the only way to remove the human element. >> Yeah, right. Tell me, who programs the computers? > Weak. Math people toss that up all the time like because computer > programmers can make mistakes in the programs computer checking of > mathematical arguments claimed to be proofs can't work. > Trouble is, even AFTER a computer checks, a human being can check as > well. > And the computer will make less errors as time goes on and the programs > are steadily improved upon, until it is perfect. > As an example, consider compilers. They are in many ways checkers, and > can catch a lot of programming errors, though not all, and over time > they've gotten better and better. > A mathematical proof is kind of like a computer program in that it goes > step by step, but perfectly, with absolutely no errors. > A proof is a claim of having a mathematical proof--a claim of > absolute perfection. > Computers CAN be perfect, but human beings just keep screwing things > up. > It's a safer bet on computers checking mathematical arguments plus > human beings, if needed, versus just letting some dude eye-ball some > arguments and claim that they are correct. Summary: mathematician discovered flaw in processor arithmetic Thomas Nicely discovered a bug in one of the early Pentium (TM) family of processors. His main research interests center on gaps in the primes, the estimation of Brun's constant, and other problems in computational number theory. David Bernier [Chris Caldwell's ``Prime pages]: http://primes.utm.edu/glossary/page.php?sort=BrunsConstant [Thomas Nicely's home page]: http://www.trnicely.net/ (see in particular the `` P*NTIUM FDIV FLAW'' section) [message posted to Usenet by Andy Grove]: > James Harris === Subject: Re: Surrogate factoring, dead idea > Computer checking is the only way to remove the human element. > Yeah, right. Tell me, who programs the computers? > Weak. Math people toss that up all the time like because computer > programmers can make mistakes in the programs computer checking of > mathematical arguments claimed to be proofs can't work. > Trouble is, even AFTER a computer checks, a human being can check as > well. > And the computer will make less errors as time goes on and the programs > are steadily improved upon, until it is perfect. Have you ever heard of the phrase Garbage In, Garbage Out? If a computer is given a bad program to execute, it will dutifully execute it. Strike 1. > As an example, consider compilers. They are in many ways checkers, and > can catch a lot of programming errors, though not all, and over time > they've gotten better and better. They are _translators_; they translate a higher-leveled language into a lower-level language, from one humans can deal with to ones computers can deal with. And they are only good at syntax. A compiler will catch the error for (i=1, i <= 10, i++) { ... } but not for (i=1; i <= 10; j++) { ... } which (if i is not changed inside the loop) will result in an infinte loop). Strike 2. > A mathematical proof is kind of like a computer program in that it goes > step by step, but perfectly, with absolutely no errors. > A proof is a claim of having a mathematical proof--a claim of > absolute perfection. So which did you mean when you said you had a proof of some result, a few months back? Did you have (a) a mathematical proof, or (b) a claim that you had a mathematical proof? There _is_ a difference between the two. I can claim that 2 + 2 = 5, but that doesn't mean that I actually have a proof that says so. > Computers CAN be perfect, but human beings just keep screwing things > up. > It's a safer bet on computers checking mathematical arguments plus > human beings, if needed, versus just letting some dude eye-ball some > arguments and claim that they are correct. Computers are good at computation (hence the name). If you want a computer to check a proof, you need to write a program which reduces a proof to something it can deal with. --- Christopher Heckman === Subject: Re: Surrogate factoring, dead idea > They are _translators_; they translate a higher-leveled language into a > lower-level language, from one humans can deal with to ones computers > can deal with. And they are only good at syntax. A compiler will catch > the error > for (i=1, i <= 10, i++) { ... } > but not > for (i=1; i <= 10; j++) { ... } > which (if i is not changed inside the loop) will result in an infinte > loop). > Strike 2. But good compilers can nowadays give hints that there may be something wrong with your program logic. Delphi is such a compiler. - If you e.g. declare a variable, but never use it, Delphi will hint you about this fact, and can thus point out the fact that you may be using another variable, where you should have used the declared-but-now-never-used one instead. - It e.g. also warns you about uninitialized variables that you nevertheless use in e.g. an expression. - If you override a method, but forget to mark it as such (a common mistake), your program might end up calling the wrong virtual method. Delphi dutifully warns you of this fact. These are just some examples of how intelligent these mere translators (as you call them) have become. This intelligence is there to prevent bugs, and it actually does work. M. === Subject: Re: Surrogate factoring, dead idea <5cVig.370808$xt.145993@fe3.news.blueyonder.co.uk They are _translators_; they translate a higher-leveled language into a > lower-level language, from one humans can deal with to ones computers > can deal with. And they are only good at syntax. A compiler will catch > the error > for (i=1, i <= 10, i++) { ... } > but not > for (i=1; i <= 10; j++) { ... } > which (if i is not changed inside the loop) will result in an infinte > loop). > Strike 2. > But good compilers can nowadays give hints that there may be something > wrong with your program logic. Delphi is such a compiler. > - If you e.g. declare a variable, but never use it, Delphi will hint you > about this fact, and can thus point out the fact that you may be using > another variable, where you should have used the > declared-but-now-never-used one instead. > - It e.g. also warns you about uninitialized variables that you > nevertheless use in e.g. an expression. gcc will also catch these two things as well. > - If you override a method, but forget to mark it as such (a common > mistake), your program might end up calling the wrong virtual method. > Delphi dutifully warns you of this fact. > These are just some examples of how intelligent these mere translators > (as you call them) have become. This intelligence is there to prevent > bugs, and it actually does work. But these checks are really secondary. My point was there is no AI built into them. --- Christopher Heckman === Subject: Re: Surrogate factoring, dead idea up. they cannot be perfect, read up on reliability theory and never use the word perfect. === Subject: Re: Surrogate factoring, dead idea <448b5e9e$0$76118$892e7fe2@authen.yellow.readfreenews.net up. > they cannot be perfect, read up on reliability theory > and > never use the word perfect. I like the word perfect. And computers can be functionally perfect in that you can wait till you die, humanity goes extinct, and well, lots of time goes by before a system makes a mistake. Sure, in infinity, or over billions of years, yeah, who could make a system that perfect? But who needs it either? A system can be built that never makes a mistake within the time-frame relevant to a human being. Sure, maybe for a god it might not work as well, but none of you are gods. James Harris === Subject: Re: Surrogate factoring, dead idea >> Computers CAN be perfect, but human beings just keep screwing things >> up. >> they cannot be perfect, read up on reliability theory >> and >> never use the word perfect. > I like the word perfect. And computers can be functionally perfect in > that you can wait till you die, humanity goes extinct, and well, lots > of time goes by before a system makes a mistake. actually not, a computer only lasts about 4 years then it is obsolete, also the storage media is changing so fast that it is not good either, one drop of catsup on the back of a CD ruins it by eating a hole in the Al where the data was stored. It is a real problem. What do you store the data on? magnetic tape? crumbles with time, CD? acid and humidity and O2 all attack it. paper tape? Google ? no retention is not through, or guaranteed === Subject: Re: Surrogate factoring, dead idea >> Computers CAN be perfect, but human beings just keep screwing things >> up. >> they cannot be perfect, read up on reliability theory >> and >> never use the word perfect. > I like the word perfect. And computers can be functionally perfect in > that you can wait till you die, humanity goes extinct, and well, lots > of time goes by before a system makes a mistake. > actually not, > a computer only lasts about 4 years then it is obsolete, also the storage > media is changing so fast that it is not good either, one drop of catsup on > the back of a CD ruins it by eating a hole in the Al where the data was > stored. > It is a real problem. > What do you store the data on? magnetic tape? crumbles with time, > CD? acid and humidity and O2 all attack it. > paper tape? > Google ? no retention is not through, or guaranteed You can post it to sci.math along with a few paranoid rampblings and people will archive it all over the net. === Subject: Re: Surrogate factoring, dead idea >It is a real problem. >What do you store the data on? magnetic tape? crumbles with time, >CD? acid and humidity and O2 all attack it. >paper tape? So far the best practical, low cost, long term storage I've heard about is printing out machine readable bar codes on acid-free paper with acid free ink to be stored in a climate controlled space, estimated time of survival of the medium is about 500 years. If money is no object feel free to etch on titanium plates like the $cientologi$ts did with their L. Ron Hubbard writings. :) === Subject: Re: Surrogate factoring, dead idea > >Computers CAN be perfect, but human beings just keep screwing things > up. > they cannot be perfect, read up on reliability theory > and > never use the word perfect. >> I like the word perfect. And computers can be functionally perfect in >> that you can wait till you die, humanity goes extinct, and well, lots >> of time goes by before a system makes a mistake. > actually not, > a computer only lasts about 4 years then it is obsolete, also the storage > media is changing so fast that it is not good either, one drop of catsup > on the back of a CD ruins it by eating a hole in the Al where the data was > stored. > It is a real problem. > What do you store the data on? magnetic tape? crumbles with time, > CD? acid and humidity and O2 all attack it. > paper tape? > Google ? no retention is not through, or guaranteed forgot, radiation and cosmic rays cause errors in silicon memory, real problem in outer space, and is a source of computer errors on earth surface too. === Subject: SF: More equations to check, no claims Well, as usual, after giving up on surrogate factoring, I thought I'd try just one more idea: T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) where 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 and I used the identity x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z) with the condition that 2xy = 2z - 1 so that I could get something weird, hoping it might work. Oddly enough, in general, it'll give non-rational x and y, so if it did work, it'd be a solution using radicals that behave just enough where it counts. Just thought up this latest. Really just doodling. Don't want to let SF die. James Harris === Subject: Re: SF: More equations to check, no claims > Well, as usual, after giving up on surrogate factoring, I thought I'd > try just one more idea: > T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) Damn typo. > where > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 > and I used the identity > x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z) > with the condition that > 2xy = 2z - 1 > so that I could get something weird, hoping it might work. > Oddly enough, in general, it'll give non-rational x and y, so if it did > work, it'd be a solution using radicals that behave just enough where > it counts. Hmmm...could this damn thing actually work? > Just thought up this latest. Really just doodling. Don't want to let > SF die. Yeah, just thought it up just a few minutes ago and dumped it out. James === Subject: Re: SF: More equations to check, no claims > Well, as usual, after giving up on surrogate factoring, I thought I'd > try just one more idea: > T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) > T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) > Damn typo. > where > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 Should be 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 which was just a mistake. So it's T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) and 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 and I wonder if that works? It might, but I still can't believe it. > and I used the identity > x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z) > with the condition that > 2xy = 2z - 1 > so that I could get something weird, hoping it might work. The weird thing I might have managed is using hyperbolas for the factors of the target. Just doodling. Wouldn't that be an odd ending? But it still doesn't feel like a solution. And it's so damn ugly. James Harris === Subject: Re: SF: More equations to check, no claims > Well, as usual, after giving up on surrogate factoring, I thought I'd > try just one more idea: > > T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) > T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) > Damn typo. > > where > > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 > Should be > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 > which was just a mistake. > So it's > T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) > and > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 > and I wonder if that works? It might, but I still can't believe it. > > and I used the identity > > x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z) > > with the condition that > > 2xy = 2z - 1 > > so that I could get something weird, hoping it might work. > The weird thing I might have managed is using hyperbolas for the > factors of the target. > Just doodling. Wouldn't that be an odd ending? But it still doesn't > feel like a solution. It's not. y^2 is usually NOT rational. > And it's so damn ugly. Yup. James Harris === Subject: Re: SF: More equations to check, no claims >> Well, as usual, after giving up on surrogate factoring, I thought I'd >> try just one more idea: >> T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) >> T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) >> Damn typo. >> where >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 >> Should be >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 >> which was just a mistake. >> So it's >> T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) >> and >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 >> and I wonder if that works? It might, but I still can't believe it. >> and I used the identity >> x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z) >> with the condition that >> 2xy = 2z - 1 >> so that I could get something weird, hoping it might work. >> The weird thing I might have managed is using hyperbolas for the >> factors of the target. >> Just doodling. Wouldn't that be an odd ending? But it still doesn't >> feel like a solution. > It's not. y^2 is usually NOT rational. sure it is. >> And it's so damn ugly. > Yup. JSH never said Yup before. What is going on? === Subject: Re: SF: More equations to check, no claims >> Yup. > JSH never said Yup before. What is going on? Sure about that? Look here: Jose Carlos Santos === Subject: Re: SF: More equations to check, no claims > JSH never said Yup before. What is going on? I think I've seen yups. === Subject: Re: SF: More equations to check, no claims <448b850a$0$15725$892e7fe2@authen.yellow.readfreenews.net> Well, as usual, after giving up on surrogate factoring, I thought I'd >> try just one more idea: >> T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) >> T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) >> Damn typo. >> where >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 >> Should be >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 >> which was just a mistake. >> So it's >> T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) >> and >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 >> and I wonder if that works? It might, but I still can't believe it. >> and I used the identity >> x^2 + 2xy + y^2 = z^2 + (x+y-z)(x+y+z) >> with the condition that >> 2xy = 2z - 1 >> so that I could get something weird, hoping it might work. >> The weird thing I might have managed is using hyperbolas for the >> factors of the target. >> Just doodling. Wouldn't that be an odd ending? But it still doesn't >> feel like a solution. > It's not. y^2 is usually NOT rational. > sure it is. >> And it's so damn ugly. > Yup. > JSH never said Yup before. What is going on? Time to check for pods. Especially if he apologizes for insisting that SF works. --- Christopher Heckman === Subject: Re: SF: More equations to check, no claims >Well, as usual, after giving up on surrogate factoring, I thought I'd >try just one more idea: >T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) >> T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) >>Damn typo. >where >16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 > Should be > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 > which was just a mistake. > So it's > T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) > and > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 > and I wonder if that works? It might, but I still can't believe it. Mathematica says you erred. Your second equation should be 16(2(y^2 - 1)*x - y)^2 - 16y^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2. With that change, you do get your first equation, FWIW. Rick === Subject: Re: SF: More equations to check, no claims >> Well, as usual, after giving up on surrogate factoring, I thought I'd >> try just one more idea: >> T = (x + y - (2x+1)/2)*(x + y + (2xy+1)/2) >> T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) >> Damn typo. >> where >> 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T + 5)^2 - (4T - 3)^2 > Should be > 16(2(y^2 - 1)*x - y)^2 = (8y^2 - 4T - 5)^2 - (4T - 3)^2 > which was just a mistake. > So it's > T = (x + y - (2xy+1)/2)*(x + y + (2xy+1)/2) nope, that is wrong too, should be T = ((x + y - ( 2 * x * y + 1) / 2 ) * ( x + y + ( 2 * x * y + 1) / 2 )) === Subject: Re: SF: More equations to check, no claims > nope, that is wrong too, should be > T = ((x + y - ( 2 * x * y + 1) / 2 ) * ( x + y + ( 2 * x * y + 1) / 2 )) I'm glad you simplified it... no z variable? Things are looking encouraging! Keep up the GOOD WORK. === Subject: Re: SF: More equations to check, no claims >> nope, that is wrong too, should be >> T = ((x + y - ( 2 * x * y + 1) / 2 ) * ( x + y + ( 2 * x * y + 1) / 2 )) > I'm glad you simplified it... no z variable? Things are looking > encouraging! Keep up the GOOD WORK. and T = (x+y)^2 - ((2xy+1)^2)/4, ............... now what ? === Subject: Re: SF: More equations to check, no claims <448b635c$0$51839$892e7fe2@authen.yellow.readfreenews.net> <448cc2b8$0$15331$892e7fe2@authen.yellow.readfreenews.net> nope, that is wrong too, should be >> T = ((x + y - ( 2 * x * y + 1) / 2 ) * ( x + y + ( 2 * x * y + 1) / 2 )) > I'm glad you simplified it... no z variable? Things are looking > encouraging! Keep up the GOOD WORK. > and T = (x+y)^2 - ((2xy+1)^2)/4, ............... now what ? Nothing. The idea fails as a practical way to factor. Just some doodling on my part that goes nowhere. James Harris === Subject: Re: SF: More equations to check, no claims : Just thought up this latest. Really just doodling. Don't want to let : SF die. Let it die. Besides, there are parabolas waiting to be rotated into hyperbolas! Justin === Subject: Re: SF might work, bad luck? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >> Like, rotate a parabola 45 degrees. >y = x^2 [1] >Let us rotate it 45 degress clockwise. You were supposed to rotate it towards you. === Subject: Re: SF might work, bad luck? Richard Tobin a .8ecrit : > Like, rotate a parabola 45 degrees. >> y = x^2 [1] >> Let us rotate it 45 degress clockwise. > You were supposed to rotate it towards you. This still get a parabola , almost obviously if you realize that you are just cutting a parabolic cylinder by a plane. > === Subject: Re: SF might work, bad luck? >As a hobby when I was a kid I was fascinated with parabolas. >>I'd draw >>them, play with the equations for them. Trace out tangent >>lines, and >>marvel over their properties--and I liked to rotate them into >>hyperbolas. Just for fun. >>Rotate it all you will, you'll never make a parabola into a >hyperbola. >>Parabola: y = ax^2 + bx + c >Hyperbola: (x-a)(y-b) = c >> Really? Are you sure? >> ___JSH >> The parabola is connected. The hyperbola is not connected. >> There is no continuous transformation of the plane that turns >> a parabola into a hyperbola. Even if you only consider one >> of the components of the hyperbola, a rotation of the plane >> will not yield a hyperbolic curve. >> This can be done in the projective plane, if you push the >> parabola through the line at infinity. The parabola is >> tangent to that line, and if it is pushed in one direction >> you get an ellipse; if you push in the opposite direction, >> you get a hyperbola. >> Dale. > Try it. Rotate a parabola and see what you get. > If it seems too abstract, post what you get. > Like, rotate a parabola 45 degrees. > Post the result. You are right, James. I checked it with Mathematica: In[1]:= y = x^2 Out[1]= y = x^2 In[2]:= CurveName[y] Out[2]= 'Parabola' In[3]:= Rotate[y, 45 Degree] Out[3]= Ok In[4]:= CurveName[y] Out[4]= 'Hyperbola' -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: JSH: Hyperbolas, parabolas Oh yeah I made a recent mistake where I talked about rotating parabolas into hyperbolas, which it was pointed out is wrong. Well, I was a kid when I was bothering with doing such things, and I had weird things that bugged me, like when you rotate a parabola, you get two curves, like the damn thing reflects around something, which is like a hyperbola, where you have two curves. Being a kid, I supposed that the parabola had been turned into a hyperbola, and didn't think about it much thereafter and just tossed that out as an adult, though considering now, yeah, that's wrong. I would play with the equations and play with them trying to force the damn parabola to just be one after the rotation, but I'd always get two curves. Oh, thinking of hyperbolas, I had another surrogate factoring idea, which boils down to using two hyperbolas for the factors of your target, which is even weirder in that x and y of the hyperbolas will tend to be non-rational but xy and x+y will not. If it works it will be one of the ugliest solutions I've ever come up with. It looks to me like hey, it might work, but I've thought that before, so I also think, hey, no way it works. Hyperbolas and parabolas. Thing is, I only liked parabolas as a kid. I'd draw them and draw them. And rotate them, and wonder why I'd get two. But I rarely bothered with any other conic section. And now here I am with freaking hyperbolas trying to factor with the damn things. James Harris === Subject: Re: JSH: Hyperbolas, parabolas > Well, I was a kid when I was bothering with doing such things, and I > had weird things that bugged me, like when you rotate a parabola, you > get two curves, like the damn thing reflects around something, which is > like a hyperbola, where you have two curves. Huh? I have a feeling you're using the word rotate to mean something it doesn't mean to other people. What do you think you get if you rotate, say, a square 45 degrees? > James Harris === Subject: Re: JSH: Hyperbolas, parabolas : Being a kid, I supposed that the parabola had been turned into a : hyperbola, and didn't think about it much thereafter and just tossed : that out as an adult, though considering now, yeah, that's wrong. : I would play with the equations and play with them trying to force the : damn parabola to just be one after the rotation, but I'd always get two : curves. You see, this is exactly how you've done all your other mathematics, too. Something looks like it might make sense or give a result you'd like and so you keep hammering away uselessly. When you're a kid this is acceptable but not when you're an adult interacting with other adults. : Oh, thinking of hyperbolas, I had another surrogate factoring idea, : which boils down to using two hyperbolas for the factors of your : target, which is even weirder in that x and y of the hyperbolas will : tend to be non-rational but xy and x+y will not. Aw, crap, a non-rational hyperbola...er...whatever that is. : And now here I am with freaking hyperbolas trying to factor with the : damn things. Oh you're so schmart. Justin === Subject: Re: JSH: Hyperbolas, parabolas > Oh yeah I made a recent mistake where > I talked about rotating parabolas into > hyperbolas, which it was pointed out is wrong. LOL. === Subject: Re: JSH: Hyperbolas, parabolas > Oh yeah I made a recent mistake where I talked about rotating parabolas > into hyperbolas, which it was pointed out is wrong. > Well, I was a kid when I was bothering with doing such things, and I > had weird things that bugged me, like when you rotate a parabola, you > get two curves, like the damn thing reflects around something, which is > like a hyperbola, where you have two curves. A rotation of the plane is a continuous function, mapping the plane back into itself. As a continuous function, it cannot turn a connected set (such as a parabola) into a set that is not connected (such as a hyperbola). You may have been doing something else, may have made errors, all sorts of explanations can be guessed at, and none are of any importance. > Being a kid, I supposed that the parabola had been turned into a > hyperbola, and didn't think about it much thereafter and just tossed > that out as an adult, though considering now, yeah, that's wrong. > I would play with the equations and play with them trying to force the > damn parabola to just be one after the rotation, but I'd always get two > curves. Whatever. This is the sort of thing that a knowledgeable instructor is good at helping the student make sense out of, and not only would your recollection not have contained the residual error, you would also have learned something extra in the process. ... the rest deleted ... Dale. === Subject: Re: JSH: Hyperbolas, parabolas the plane back into itself. As a continuous function, it > cannot turn a connected set (such as a parabola) into a > set that is not connected (such as a hyperbola). > You may have been doing something else, may have made > errors, all sorts of explanations can be guessed at, and > none are of any importance. A hyperbola and a parabola are birationally equivalent, and you can cook up a family of rational maps which send hyperbolas to hyperbolas and in the limit reach a parabola; you can think of this as fixing one focus and moving the other until it reaches the line at infinity. === Subject: Re: JSH: Hyperbolas, parabolas into hyperbolas, which it was pointed out is wrong. > Well, I was a kid when I was bothering with doing such things, and I > had weird things that bugged me, like when you rotate a parabola, you > get two curves, like the damn thing reflects around something, which is > like a hyperbola, where you have two curves. > A rotation of the plane is a continuous function, mapping > the plane back into itself. As a continuous function, it > cannot turn a connected set (such as a parabola) into a > set that is not connected (such as a hyperbola). > You may have been doing something else, may have made > errors, all sorts of explanations can be guessed at, and > none are of any importance. Maybe he means he rotated it, and took the union of the two curves (old and new). --- Christopher Heckman === Subject: Re: JSH: Hyperbolas, parabolas Proginoskes a .8ecrit : > Oh yeah I made a recent mistake where I talked about rotating parabolas > into hyperbolas, which it was pointed out is wrong. > Well, I was a kid when I was bothering with doing such things, and I > had weird things that bugged me, like when you rotate a parabola, you > get two curves, like the damn thing reflects around something, which is > like a hyperbola, where you have two curves. >> A rotation of the plane is a continuous function, mapping >> the plane back into itself. As a continuous function, it >> cannot turn a connected set (such as a parabola) into a >> set that is not connected (such as a hyperbola). >> You may have been doing something else, may have made >> errors, all sorts of explanations can be guessed at, and >> none are of any importance. > Maybe he means he rotated it, and took the union of the two curves (old > and new). If it is of any importance, i believe he discovered that after any rotation it becomes impossible to express the equation as a unique function (for instance, y=x^2 becomes y=+/- sqrt(x) after a rotation of 90Á) > --- Christopher Heckman === Subject: Re: JSH: Hyperbolas, parabolas <448bb2b6$0$7764$7a628cd7@news.club-internet.fr Proginoskes a .8ecrit : > Oh yeah I made a recent mistake where I talked about rotating parabolas > into hyperbolas, which it was pointed out is wrong. >> Well, I was a kid when I was bothering with doing such things, and I > had weird things that bugged me, like when you rotate a parabola, you > get two curves, like the damn thing reflects around something, which is > like a hyperbola, where you have two curves. > A rotation of the plane is a continuous function, mapping >> the plane back into itself. As a continuous function, it >> cannot turn a connected set (such as a parabola) into a >> set that is not connected (such as a hyperbola). >> You may have been doing something else, may have made >> errors, all sorts of explanations can be guessed at, and >> none are of any importance. > Maybe he means he rotated it, and took the union of the two curves (old > and new). > If it is of any importance, i believe he discovered that after any > rotation it becomes impossible to express the equation as a unique > function (for instance, y=x^2 becomes y=+/- sqrt(x) after a rotation of 90Á) Hey, that must be it! It bugged me as a kid. It seemed, wrong, as I started with one curve, did a rotation, and ended up with two! It really bugged me for a while. It just didn't seem right. James Harris === Subject: Re: JSH: Hyperbolas, parabolas <448bb2b6$0$7764$7a628cd7@news.club-internet.fr Proginoskes a .8ecrit : > Oh yeah I made a recent mistake where I talked about rotating parabolas > into hyperbolas, which it was pointed out is wrong. >> Well, I was a kid when I was bothering with doing such things, and I > had weird things that bugged me, like when you rotate a parabola, you > get two curves, like the damn thing reflects around something, which is > like a hyperbola, where you have two curves. >>> A rotation of the plane is a continuous function, mapping >> the plane back into itself. As a continuous function, it >> cannot turn a connected set (such as a parabola) into a >> set that is not connected (such as a hyperbola). >> You may have been doing something else, may have made >> errors, all sorts of explanations can be guessed at, and >> none are of any importance. > > Maybe he means he rotated it, and took the union of the two curves (old > and new). > If it is of any importance, i believe he discovered that after any > rotation it becomes impossible to express the equation as a unique > function (for instance, y=x^2 becomes y=+/- sqrt(x) after a rotation of 90Á) > Hey, that must be it! It bugged me as a kid. > It seemed, wrong, as I started with one curve, did a rotation, and > ended up with two! But y = +/- sqrt(x) still describes only one continuous connected curve. > It really bugged me for a while. It just didn't seem right. Did it occur to you also that a hyperbola has two straight-line asymptotes, whereas a parabola has none? And that the property of being a straight-line asymptote would be preserved by any rotation? Marcus. > James Harris === Subject: Re: JSH: Hyperbolas, parabolas > Proginoskes a .8ecrit : >> Oh yeah I made a recent mistake where I talked about rotating parabolas >> into hyperbolas, which it was pointed out is wrong. >> Well, I was a kid when I was bothering with doing such things, and I >> had weird things that bugged me, like when you rotate a parabola, you >> get two curves, like the damn thing reflects around something, which is >> like a hyperbola, where you have two curves. > A rotation of the plane is a continuous function, mapping > the plane back into itself. As a continuous function, it > cannot turn a connected set (such as a parabola) into a > set that is not connected (such as a hyperbola). > You may have been doing something else, may have made > errors, all sorts of explanations can be guessed at, and > none are of any importance. >> Maybe he means he rotated it, and took the union of the two curves (old >> and new). > If it is of any importance, i believe he discovered that after any > rotation it becomes impossible to express the equation as a unique > function (for instance, y=x^2 becomes y=+/- sqrt(x) after a rotation of > 90Á) some functions would survive, most linear, monotonically increasing/decreasing, ones were the slope never has a value of 0, This is a great personal discovery for some. === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at [...] > And saying that my proof is invalid because I never considered the > possibility of using math techniques such as mathematical induction and > proof by contradiction is also on par with this type of argument that > the Pythagorean Theorem is wrong. No, a better analogy would be that your argument is like the proof that it is impossible to square the circle or trisect an angle. In both cases, they are proofs that there is no way to do something. [back up in the post:] > The only axiom needed for my proof to work is the axiom that in order > to prove that T^k(n)=1, it is necessary to specify the formula for > T^k(n) in the proof. > An axiom is an assumption, and you haven't stated this in your paper. > If you had stated: > There is no proof of T^k(n) = 1 which includes a formula for T^k(n) in > it. > It is obviously true that in order to prove that T^k(n)=1, it is > necessary to specify the formula for T^k(n) in the proof. I'll give > you an example: Let's say you are a math teacher and you give your > class a homework assignment to prove that g(n)=1. The most likely > reaction of the class would be to protest by saying you have to tell us > what the formula for g(n) is. Yes, I need to give a description of g(n), but that description doesn't need to be an explicit formula. For instance, if I said that g(0) = 0, g(1) = 1, and g(n) = g(n-1) + g(n-2) if n >= 2, then no one would complain that they can't calculate g(10). --- Christopher Heckman === Subject: Re: Another Reason Why Collatz is Unprovable # # I think the context of my argument that Collatz is unprovable is clear # - What I mean by unprovable is unprovable in any reasonable axiom # system. This would exclude inconsistent axiom systems. The fact that I # never specified the axiom system does not take away from my argument. Don't be ridiculous. Theorem 3 of your paper The Collatz 3n+1 Conjecture is Unprovable reads as follows: It is impossible to prove the 3n+1 Conjecture. Now suppose that the 3n+1 Conjecture is true. Consider PA plus the axiom that the Collatz algorithm always reaches 1. In this axiom system, there is a very short and very trivial proof that the Collatz algorithm always reaches 1. But your proof states that this is absolutely impossible under any Feinstein-reasonable axiom system. Now the big question arises: If one adds a true statement as axiom to PA, is the resulting axiom system Feinstein-reasonable? --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: Re: Another Reason Why Collatz is Unprovable # #> If you ignore context, then the fact the sum of the angles of a #> triangle is 180 degrees is both provable (in Euclidean geometry) and #> unprovable (in non-Euclidean geometry). #> Now, there is nothing wrong with an axiom system where you take #> standard mathematics and add the axiom 2+2=5. It turns out that in this #> system, _every_ proposition is provable. However, the system is not #> consistent -- that is, you can prove statements which are false -- so #> it's not useful. # # I think the context of my argument that Collatz is unprovable is clear # - What I mean by unprovable is unprovable in any reasonable axiom # system. This would exclude inconsistent axiom systems. The fact that I # never specified the axiom system does not take away from my argument. This reminds me (a lot!) of the long discussion in comp.theory around your incorrect proof for the P versus NP problem: 1. There you first claimed that your proof works for ALL models of computation. 2. Later, you claimed that your proof works for ALL REASONABLE models of computation (but without being able to specify the properties of a reasonable model of computation to us). 3. In the end, you finally stated that your proof only works in the so-called Mable-Mildred-Feinstein model of computation, which has nothing to do with the P versus NP problem. --------------------------- It looks as if your Collatz paper is suffering from similar diseases. You do NOT specify your axiom system to the reader, although you make strong and very restrictive assumptions on possible proofs within your model. For instance, your Theorem 2 states: [...] then in order to prove that [...] it is necessary to specify the values [...] For Goedel, a proof was nothing but a finite sequence of characters. In the Feinstein-model-of-proof, however, a proof is something THAT SPECIFIES CERTAIN VALUES. --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at It is obviously true that in order to prove that T^k(n)=1, it is > necessary to specify the formula for T^k(n) in the proof. You are quite wrong about this. In fact, what you need to prove is not that T^k(n) = 1, but that for every n there exists a k such that T^k(n) = 1. A general existence proof need not exhibit an actual value of k for anything other than a finite set of n (if that) -- and it certainly does not need to include a general formula for T^k(n). === Subject: Re: Another Reason Why Collatz is Unprovable [...] > The only axiom needed for my proof to work is the axiom that in order > to prove that T^k(n)=1, it is necessary to specify the formula for > T^k(n) in the proof. I don't know anyone who doesn't accept this axiom > as obviously true. What you are still missing is that we may specify by rule for all integers, not by roster for each individually. Many people have now tried to explain this simple concept to you in many ways, but you still haven't gotten it. Learning the basics of mathematic proof will serve you much better than charging unarmed into problems like Collatz and P?=NP. -- --Bryan === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at [...] > If you > think mathematics is some kind of game, then this is perfectly OK, but > if you believe like me that mathematics is more than just a game, then > your argument is very weak. > I, for one, take mathematics seriously. > The only axiom needed for my proof to work is the axiom that in order > to prove that T^k(n)=1, it is necessary to specify the formula for > T^k(n) in the proof. > An axiom is an assumption, and you haven't stated this in your paper. > If you had stated: > There is no proof of T^k(n) = 1 which includes a formula for T^k(n) in > it. > It is obviously true that in order to prove that T^k(n)=1, it is > necessary to specify the formula for T^k(n) in the proof. I'll give > you an example: Let's say you are a math teacher and you give your > class a homework assignment to prove that g(n)=1. The most likely > reaction of the class would be to protest by saying you have to tell us > what the formula for g(n) is. It's the same thing with the argument of > my paper. The argument of your paper is about proving a related result: That g(n) = 1 for all n. Your paper says that basically, you must show that g(1) = 1, g(2) = 1, g(3) = 1, etc. and that there are no other proof systems. > Then I would agree with this. But to prove that Collatz is unprovable, > you need to _prove_ your axiom, not just _state_ it. > For instance, the following is true and provable: > There is no proof of T^k(n) = 1 if 2 = 0.5 > (Proof: Suppose n > 0. If n is odd, then T(n) = 3n+1 > n; if n is even, > T(n) = n/2 = n/0.5 = 2n > n. Thus T^k(n) >= n + k for all nonnegative > k, and so is never 1.) > But it doesn't count as a proof of Collatz, because I would need to > justify the axiom > 2 = 0.5. > Another example is Euclidean versus non-Euclidean geometry that I > mentioned above. > I don't know anyone who doesn't accept this axiom as obviously true. > And the reason I don't accept the axiom is because of the principle of > generalization: > Suppose you have a proof that object X has property P. If every step of > the proof is valid for the object Y, then Y has property P (and this is > provable). > Once again, let U be the modified Collatz function > U(n) = n/2, if n is even, > U(n) = (n-1)/2, if n is odd. > I now claim that your proof can be used, step by step, to prove that > the statement U(n) is eventually 0 is unprovable, using the principle > of generalization above. > Your Principle of Generalization works only if mathematical proofs were > written, read, and checked by computers. (1) So that means your paper can't be checked by computer? How do you know there isn't some subtle flaw then? (2) The principle of generalization _does_ work for mathematical proofs written by human beings. (3) You would accept it if it weren't being used to question your proof. In fact, in your proof, you supply a formula for T^k(n) that holds for an infinite number of values of n --- and prove that it does --- instead of considering them all one by one. Hence, you are using the Principle of Generalization in that result. (Or Lagarias is.) > But since they are not, your > Principle of Generalization only works in theory. In practice, since > proofs are as of today made by human beings, there is a such thing as > interpretation and reading between the lines. Not everything is always > spelled out. > In this case, it is implicit in the argument that there is a one-to-one > correspondence between all of the possible formulas for T^k and all > possible values of (n,T(n),...,T^(k-1)(n)) mod 2 that the possible > formulas for T^k cannot be reduced any further for any k. In the case > of your function U, which is really the greatest integer function, the > possible formulas of U^k can be reduced to zero when k is on the order > of log n. So your argument is: My proof doesn't apply to U, because _I_ can see how to simplify U^k(n). In other words, a proof from ignorance. (_I_ don't see how to reduce T^k(n), so Collatz must be unprovable.) But your _proof_ doesn't make this distinction. And it's your proof that's in debate, not your personal skills. > Another reason the axiom is questionable is because a proof could > proceed by establishing that > (1) T^k(n) <= F(k,n), for some formula F (maybe for big enough n and k), > (2) limsup F(k,n) <= C for some small constant C (where the limit is > as k goes to infinity). > Part (1) could be established without an explicit formula for T^k(n) > (by induction, for instance). > The only problem is that proving such would not prove Collatz, that > T^k(n)=1. It's still conceivable that there exist nontrivial cycles > (not 1,2,1,2,...) of T^k(n) bounded above by C for an infinite number > of numbers n. If C = 4, then it will prove Collatz. If C <= 1000, it will prove Collatz. That's what I mean by small: small enough so that the rest can be done by brute force. --- Christopher Heckman === Subject: Re: Another Reason Why Collatz is Unprovable Am 10.06.2006 02:09 schrieb Proginoskes: > (by induction, for instance). >> The only problem is that proving such would not prove Collatz, that >> T^k(n)=1. It's still conceivable that there exist nontrivial cycles >> (not 1,2,1,2,...) of T^k(n) bounded above by C for an infinite number >> of numbers n. > If C = 4, then it will prove Collatz. If C <= 1000, it will prove > Collatz. That's what I mean by small: small enough so that the rest > can be done by brute force. > --- Christopher Heckman There are already proofs, that certain type of loops cannot exist. These are the so-called 1-cycles, 2-cycles,... m-cycles (m up to 68) where 1-cycle means here, that the sequence of values along the transformation only ascends and then only descends, like a hill with one peak, 2-cycles two peaks and so on. The first proof against an 1-cycle is due to Ray Steiner, 1978, and the second about 3-cycles up to 68-cycles are due to Benne de Weger and John Simons, using the Steiner- approach (the B de Weger paper is/was online). The notion of cycle includes, that the number of ascending and descending steps is not bounded, so for any finite length the *type* of 1-cycle, 2-cycle etc are disproven. Also it can be shown, that the construction of a loop (any loop, not only 1-peak,2-peak.,,, cycles) of a *certain length*, but of arbitrary peaks-structure, restricts the values for the involved numbers with a *high* bound. With that criterion I can disprove many cycle lengthes up to ~ 200 with completely elementary means. (a sketch of this is at http://www.uni-kassel.de/fg_pur/helms/math/collatz/aboutloop/ at entry the general loop) Gottfried Helms === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at > #> My proof on arXiv.org rules out a proof by contradiction because it > #> rules out any proof. It is a completely logical proof. No one on usenet > #> has found any flaws in the proof. The only things people have been > #> doing are claiming that my definition of random is not rigorous > #> (because it does not specify a formal language, which is really > #> irrelevant in the context of my proof) and giving straw-man arguments > #> which prove false statements and claiming that my proof uses the same > #> type of arguments. If you have faith in logic, then you should have > #> faith in my proof. > # > # My proof never claims that you have to treat every integer > # individually. My proof says: let's pretend that we have a proof of > # Collatz with L bits. It then shows that there is a specific n for which > # any proof that Collatz halts at one with input n requires at least L+1 > # least L+1 bits, so we have a contradiction, as our pretend proof only > # has L bits. Therefore, Collatz is unprovable. > > In your paper, you do not discuss at all the axiom system that > your are using. All your statements on proofs are absolute > statements, that do not even touch the question of the underlying > axiom systems in the slightest way. > > Suppose, that I use as axiom system PA plus the axiom that the > Collatz algorithm always reaches 1. > In this axiom system, there is a very short and very trivial > proof that the Collatz algorithm always reaches 1. > This horribly collides with your absolute statement that > Collatz is unprovable. > > Using this line of reasoning, you can make 2+2=5 an axiom and then > prove 2+2=5, contradicting the well-established fact that 2+2=5 is > unprovable, i.e., that it is impossible to prove that 2+2=5. > > Whenever provability is mentioned, it has to be put in contest. The > last part of the previous sentence should read: > > ... contradicting the well-established fact that 2+2=5 is unprovable > _in the standard mathematical model_ [etc]. > > If you ignore context, then the fact the sum of the angles of a > triangle is 180 degrees is both provable (in Euclidean geometry) and > unprovable (in non-Euclidean geometry). > > Now, there is nothing wrong with an axiom system where you take > standard mathematics and add the axiom 2+2=5. It turns out that in this > system, _every_ proposition is provable. However, the system is not > consistent -- that is, you can prove statements which are false -- so > it's not useful. > I think the context of my argument that Collatz is unprovable is clear > - What I mean by unprovable is unprovable in any reasonable axiom > system. This would exclude inconsistent axiom systems. The fact that I > never specified the axiom system does not take away from my argument. > > If you > think mathematics is some kind of game, then this is perfectly OK, but > if you believe like me that mathematics is more than just a game, then > your argument is very weak. > > I, for one, take mathematics seriously. > > The only axiom needed for my proof to work is the axiom that in order > to prove that T^k(n)=1, it is necessary to specify the formula for > T^k(n) in the proof. > > An axiom is an assumption, and you haven't stated this in your paper. > If you had stated: > > There is no proof of T^k(n) = 1 which includes a formula for T^k(n) in > it. > It is obviously true that in order to prove that T^k(n)=1, it is > necessary to specify the formula for T^k(n) in the proof. I'll give > you an example: Let's say you are a math teacher and you give your > class a homework assignment to prove that g(n)=1. The most likely > reaction of the class would be to protest by saying you have to tell us > what the formula for g(n) is. It's the same thing with the argument of > my paper. > > Then I would agree with this. But to prove that Collatz is unprovable, > you need to _prove_ your axiom, not just _state_ it. > > For instance, the following is true and provable: > > There is no proof of T^k(n) = 1 if 2 = 0.5 > > (Proof: Suppose n > 0. If n is odd, then T(n) = 3n+1 > n; if n is even, > > T(n) = n/2 = n/0.5 = 2n > n. Thus T^k(n) >= n + k for all nonnegative > k, and so is never 1.) > > But it doesn't count as a proof of Collatz, because I would need to > justify the axiom > 2 = 0.5. > > Another example is Euclidean versus non-Euclidean geometry that I > mentioned above. > > I don't know anyone who doesn't accept this axiom as obviously true. > > And the reason I don't accept the axiom is because of the principle of > generalization: > > Suppose you have a proof that object X has property P. If every step of > the proof is valid for the object Y, then Y has property P (and this is > provable). > > Once again, let U be the modified Collatz function > > U(n) = n/2, if n is even, > U(n) = (n-1)/2, if n is odd. > > I now claim that your proof can be used, step by step, to prove that > the statement U(n) is eventually 0 is unprovable, using the principle > of generalization above. > Your Principle of Generalization works only if mathematical proofs were > written, read, and checked by computers. But since they are not, your > Principle of Generalization only works in theory. In practice, since > proofs are as of today made by human beings, there is a such thing as > interpretation and reading between the lines. Not everything is always > spelled out. > In this case, it is implicit in the argument that there is a one-to-one > correspondence between all of the possible formulas for T^k and all > possible values of (n,T(n),...,T^(k-1)(n)) mod 2 that the possible > formulas for T^k cannot be reduced any further for any k. In the case > of your function U, which is really the greatest integer function, the > possible formulas of U^k can be reduced to zero when k is on the order > of log n. > > Another reason the axiom is questionable is because a proof could > proceed by establishing that > > (1) T^k(n) <= F(k,n), for some formula F (maybe for big enough n and > k), > (2) limsup F(k,n) <= C for some small constant C (where the limit is > as k goes to infinity). > > Part (1) could be established without an explicit formula for T^k(n) > (by induction, for instance). > The only problem is that proving such would not prove Collatz, that > T^k(n)=1. It's still conceivable that there exist nontrivial cycles > (not 1,2,1,2,...) of T^k(n) bounded above by C for an infinite number > of numbers n. > And even if you could prove that there are no nontrivial cylces bounded > above by C, this type of argument would be on par with someone arguing > that the Pythagorean Theorem is wrong by claiming that if one were to > construct a huge right triangle with sides (a,b,c) in terms with an > area on the order of hundreds of square miles where a^2+b^2=c^2, then > this would disprove the Pythagorean Theorem and therefore the > Pythagorean Theorem is wrong because the proof of the Pythagorean > Theorem never explicitly considered this possibility. > I don't understand your point here. The Pythagorean Theorem is a > theorem (look up what that means). Collatz is still a conjecture. > It is indeed an open question whether Collatz becomes false if > the numbers are big enough. That CANNOT happen with Pythagoras > regardless of what was and was not considered. > And saying that my proof is invalid because I never considered the > possibility of using math techniques such as mathematical induction and > proof by contradiction is also on par with this type of argument that > the Pythagorean Theorem is wrong. > This analogy makes no sense. Pythagoras has been PROVED. > Collatz has not. You have to show that a proof via induction or > contradiction CANNOT exist. You have offered no such evidence. > You have shown only that a proof by YOUR methods cannot exist > and you have simply ASSUMED that proofs by other methods cannot > exist. Assuming your premise is true is NOT proof that it is true. My proof that Collatz is unprovable proves that all methods including proof by induction and proof by contradiction are useless for trying to prove the Collatz Conjecture. The people who think that my proof is invalid because it doesn't explicitly mention these types of techniques by name are talking just like the people who lived a little after Galois' time and said that his impossibility proof for finding roots of 5th degree polynomials was invalid because it didn't explicitly consider every possible formula in terms of elementary arithmetic operations for the roots of a general 5th degree polynomial, but he only considered formulas which fit into his limited model of possible formulas. The only difference between these two proofs is that my result is simple and obvious and can be easily understood with a little common sense, while Galois' proof requires a sophisticated brilliant argument which requires a lot of time to understand. It's so basic that it's amazing that there are still people out there who don't believe it - in order to know whether the Collatz algorithm halts, you have to run the algorithm. There are no shortcuts. It's really just an example of Stephen Wolfram's notion of computational irreducibility, as he talks about in his book New Kind of Science. It's just what he predicts would happen in his book when math problems reach a certain threshold of complexity and start to involve logical operations as well as equations. Once you get to that threshold, there is nothing you can do to predict what an algorithm is going to do unless you run the algorithm itself. Ever since my proof that Collatz is unprovable has been up on the internet, anyone who has read it and still thinks it is possible to prove Collatz might as well try to find a general formula for the roots of a fifth degree polynomial in terms of elementary arithmetic operations as well. Serious math journals which receive papers claiming a proof of Collatz should now begin rejecting these papers as crank papers without reading them, since such proofs are impossible. Craig Craig > --- Christopher Heckman === Subject: Re: Another Reason Why Collatz is Unprovable > My proof that Collatz is unprovable proves that all > methods including > proof by induction and proof by contradiction are > useless for trying to > prove the Collatz Conjecture. The people who think > that my proof is > invalid because it doesn't explicitly mention these > types of techniques > by name are talking just like the people who lived a > little after > Galois' time and said that his impossibility proof > for finding roots of > 5th degree polynomials was invalid because it didn't > explicitly > consider every possible formula in terms of > elementary arithmetic > operations for the roots of a general 5th degree > polynomial, but he > only considered formulas which fit into his limited > model of possible > formulas. The only difference between these two > proofs is that my > result is simple and obvious and can be easily > understood with a little > common sense, while Galois' proof requires a > sophisticated brilliant > argument which requires a lot of time to understand.<< Don't you mean Abel's Impossibility Theorem (which is consistent with Galois theory)? The depth of that result lies in demarcating arithmetic from analysis. I can't see that it has anything to do with proof strategy, as you imply. Proof methods and calculating methods are not identical. Even Chaitin, the most enamored of computing power that I know, acknowledges the role of philosophy (digital philosophy) in proofs. Do you, like Kronecker, think that arithmetic is all there is to mathematics? Tom > It's so basic that it's amazing that there are still > people out there > who don't believe it - in order to know whether the > Collatz algorithm > halts, you have to run the algorithm. There are no > shortcuts. It's > really just an example of Stephen Wolfram's notion of > computational > irreducibility, as he talks about in his book New > Kind of Science. It's > just what he predicts would happen in his book when > math problems reach > a certain threshold of complexity and start to > involve logical > operations as well as equations. Once you get to that > threshold, there > is nothing you can do to predict what an algorithm is > going to do > unless you run the algorithm itself. > Ever since my proof that Collatz is unprovable has > been up on the > internet, anyone who has read it and still thinks it > is possible to > prove Collatz might as well try to find a general > formula for the roots > of a fifth degree polynomial in terms of elementary > arithmetic > operations as well. Serious math journals which > receive papers claiming > a proof of Collatz should now begin rejecting these > papers as crank > papers without reading them, since such proofs are > impossible. > Craig === Subject: Re: Another Reason Why Collatz is Unprovable <5837210.1149938254891.JavaMail.jakarta@nitrogen.mathforum.org My proof that Collatz is unprovable proves that all > methods including > proof by induction and proof by contradiction are > useless for trying to > prove the Collatz Conjecture. The people who think > that my proof is > invalid because it doesn't explicitly mention these > types of techniques > by name are talking just like the people who lived a > little after > Galois' time and said that his impossibility proof > for finding roots of > 5th degree polynomials was invalid because it didn't > explicitly > consider every possible formula in terms of > elementary arithmetic > operations for the roots of a general 5th degree > polynomial, but he > only considered formulas which fit into his limited > model of possible > formulas. The only difference between these two > proofs is that my > result is simple and obvious and can be easily > understood with a little > common sense, while Galois' proof requires a > sophisticated brilliant > argument which requires a lot of time to understand.<< > Don't you mean Abel's Impossibility Theorem (which is > consistent with Galois theory)? http://en.wikipedia.org/wiki/Abel-Ruffini_theorem http://en.wikipedia.org/wiki/Galois_theory I didn't know that this was known before Galois. I remember learning it differently. You learn something everyday. So I take my analogy back and replace Galois' name with Ruffini, since he was first. The depth of that > result lies in demarcating arithmetic from analysis. I > can't see that it has anything to do with proof strategy, > as you imply. Proof methods and calculating methods > are not identical. Even Chaitin, the most enamored of > computing power that I know, acknowledges the role of > philosophy (digital philosophy) in proofs. > Do you, like Kronecker, think that arithmetic is all > there is to mathematics? If I had lived in Kronecker's time, I would have probably sided with him in the debate against Cantor - I don't think actual infinities have any place in mathematics. (Only potential infinities, as Aristotle distinguished them.) However, now that the genie is out of the bottle, there's nothing I can do to stop it. I would agree with Kronecker that G-d created numbers. Everything else is the work of man. Craig > Tom > It's so basic that it's amazing that there are still > people out there > who don't believe it - in order to know whether the > Collatz algorithm > halts, you have to run the algorithm. There are no > shortcuts. It's > really just an example of Stephen Wolfram's notion of > computational > irreducibility, as he talks about in his book New > Kind of Science. It's > just what he predicts would happen in his book when > math problems reach > a certain threshold of complexity and start to > involve logical > operations as well as equations. Once you get to that > threshold, there > is nothing you can do to predict what an algorithm is > going to do > unless you run the algorithm itself. > Ever since my proof that Collatz is unprovable has > been up on the > internet, anyone who has read it and still thinks it > is possible to > prove Collatz might as well try to find a general > formula for the roots > of a fifth degree polynomial in terms of elementary > arithmetic > operations as well. Serious math journals which > receive papers claiming > a proof of Collatz should now begin rejecting these > papers as crank > papers without reading them, since such proofs are > impossible. Craig === Subject: Re: Another Reason Why Collatz is Unprovable > My proof that Collatz is unprovable proves that > all > methods including > proof by induction and proof by contradiction are > useless for trying to > prove the Collatz Conjecture. The people who > think > that my proof is > invalid because it doesn't explicitly mention > these > types of techniques > by name are talking just like the people who > lived a > little after > Galois' time and said that his impossibility > proof > for finding roots of > 5th degree polynomials was invalid because it > didn't > explicitly > consider every possible formula in terms of > elementary arithmetic > operations for the roots of a general 5th degree > polynomial, but he > only considered formulas which fit into his > limited > model of possible > formulas. The only difference between these two > proofs is that my > result is simple and obvious and can be easily > understood with a little > common sense, while Galois' proof requires a > sophisticated brilliant > argument which requires a lot of time to > understand.<< > Don't you mean Abel's Impossibility Theorem (which > is > consistent with Galois theory)? > http://en.wikipedia.org/wiki/Abel-Ruffini_theorem > http://en.wikipedia.org/wiki/Galois_theory > I didn't know that this was known before Galois. I > remember learning it > differently. You learn something everyday. So I take > my analogy back > and replace Galois' name with Ruffini, since he was > first.<< Speaking for myself, I am partial to W.R. Hamilton's commentary on Abel's result. His 1839 paper to the Royal Irish Academy is very compact and eloquent, IMO. > The depth of that > result lies in demarcating arithmetic from > analysis. I > can't see that it has anything to do with proof > strategy, > as you imply. Proof methods and calculating methods > are not identical. Even Chaitin, the most enamored > of > computing power that I know, acknowledges the role > of > philosophy (digital philosophy) in proofs. > Do you, like Kronecker, think that arithmetic is > all > there is to mathematics? Feinstein: > If I had lived in Kronecker's time, I would have > probably sided with > him in the debate against Cantor - I don't think > actual infinities have > any place in mathematics. (Only potential infinities, > as Aristotle > distinguished them.) However, now that the genie is > out of the bottle, > there's nothing I can do to stop it. > I would agree with Kronecker that G-d created > numbers. Everything else > is the work of man. > Craig Does that mean that G-d didn't create analysis? :-) (Kronecker's actual statement is that G-d created the integers, or the natural numbers ...) Creation is a very risky business, though. It often misbehaves. Tom === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at My proof that Collatz is unprovable proves that all methods including > proof by induction and proof by contradiction are useless for trying to > prove the Collatz Conjecture. Errm, I don't think so. > The people who think that my proof is > invalid... Just out of interest, is there anyone who *doesn't* think your proof is invalid? [snip rest] === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at > #> My proof on arXiv.org rules out a proof by contradiction because it > #> rules out any proof. It is a completely logical proof. No one on usenet > #> has found any flaws in the proof. The only things people have been > #> doing are claiming that my definition of random is not rigorous > #> (because it does not specify a formal language, which is really > #> irrelevant in the context of my proof) and giving straw-man arguments > #> which prove false statements and claiming that my proof uses the same > #> type of arguments. If you have faith in logic, then you should have > #> faith in my proof. > # > # My proof never claims that you have to treat every integer > # individually. My proof says: let's pretend that we have a proof of > # Collatz with L bits. It then shows that there is a specific n for which > # any proof that Collatz halts at one with input n requires at least L+1 > # least L+1 bits, so we have a contradiction, as our pretend proof only > # has L bits. Therefore, Collatz is unprovable. > > In your paper, you do not discuss at all the axiom system that > your are using. All your statements on proofs are absolute > statements, that do not even touch the question of the underlying > axiom systems in the slightest way. > > Suppose, that I use as axiom system PA plus the axiom that the > Collatz algorithm always reaches 1. > In this axiom system, there is a very short and very trivial > proof that the Collatz algorithm always reaches 1. > This horribly collides with your absolute statement that > Collatz is unprovable. > > Using this line of reasoning, you can make 2+2=5 an axiom and then > prove 2+2=5, contradicting the well-established fact that 2+2=5 is > unprovable, i.e., that it is impossible to prove that 2+2=5. > > Whenever provability is mentioned, it has to be put in contest. The > last part of the previous sentence should read: > > ... contradicting the well-established fact that 2+2=5 is unprovable > _in the standard mathematical model_ [etc]. > > If you ignore context, then the fact the sum of the angles of a > triangle is 180 degrees is both provable (in Euclidean geometry) and > unprovable (in non-Euclidean geometry). > > Now, there is nothing wrong with an axiom system where you take > standard mathematics and add the axiom 2+2=5. It turns out that in this > system, _every_ proposition is provable. However, the system is not > consistent -- that is, you can prove statements which are false -- so > it's not useful. > > I think the context of my argument that Collatz is unprovable is clear > - What I mean by unprovable is unprovable in any reasonable axiom > system. This would exclude inconsistent axiom systems. The fact that I > never specified the axiom system does not take away from my argument. > > If you > think mathematics is some kind of game, then this is perfectly OK, but > if you believe like me that mathematics is more than just a game, then > your argument is very weak. > > I, for one, take mathematics seriously. > > The only axiom needed for my proof to work is the axiom that in order > to prove that T^k(n)=1, it is necessary to specify the formula for > T^k(n) in the proof. > > An axiom is an assumption, and you haven't stated this in your paper. > If you had stated: > > There is no proof of T^k(n) = 1 which includes a formula for T^k(n) in > it. > > It is obviously true that in order to prove that T^k(n)=1, it is > necessary to specify the formula for T^k(n) in the proof. I'll give > you an example: Let's say you are a math teacher and you give your > class a homework assignment to prove that g(n)=1. The most likely > reaction of the class would be to protest by saying you have to tell us > what the formula for g(n) is. It's the same thing with the argument of > my paper. > > Then I would agree with this. But to prove that Collatz is unprovable, > you need to _prove_ your axiom, not just _state_ it. > > For instance, the following is true and provable: > > There is no proof of T^k(n) = 1 if 2 = 0.5 > > (Proof: Suppose n > 0. If n is odd, then T(n) = 3n+1 > n; if n is even, > > T(n) = n/2 = n/0.5 = 2n > n. Thus T^k(n) >= n + k for all nonnegative > k, and so is never 1.) > > But it doesn't count as a proof of Collatz, because I would need to > justify the axiom > 2 = 0.5. > > Another example is Euclidean versus non-Euclidean geometry that I > mentioned above. > > I don't know anyone who doesn't accept this axiom as obviously true. > > And the reason I don't accept the axiom is because of the principle of > generalization: > > Suppose you have a proof that object X has property P. If every step of > the proof is valid for the object Y, then Y has property P (and this is > provable). > > Once again, let U be the modified Collatz function > > U(n) = n/2, if n is even, > U(n) = (n-1)/2, if n is odd. > > I now claim that your proof can be used, step by step, to prove that > the statement U(n) is eventually 0 is unprovable, using the principle > of generalization above. > > Your Principle of Generalization works only if mathematical proofs were > written, read, and checked by computers. But since they are not, your > Principle of Generalization only works in theory. In practice, since > proofs are as of today made by human beings, there is a such thing as > interpretation and reading between the lines. Not everything is always > spelled out. > > In this case, it is implicit in the argument that there is a one-to-one > correspondence between all of the possible formulas for T^k and all > possible values of (n,T(n),...,T^(k-1)(n)) mod 2 that the possible > formulas for T^k cannot be reduced any further for any k. In the case > of your function U, which is really the greatest integer function, the > possible formulas of U^k can be reduced to zero when k is on the order > of log n. > > Another reason the axiom is questionable is because a proof could > proceed by establishing that > > (1) T^k(n) <= F(k,n), for some formula F (maybe for big enough n and > k), > (2) limsup F(k,n) <= C for some small constant C (where the limit is > as k goes to infinity). > > Part (1) could be established without an explicit formula for T^k(n) > (by induction, for instance). > > The only problem is that proving such would not prove Collatz, that > T^k(n)=1. It's still conceivable that there exist nontrivial cycles > (not 1,2,1,2,...) of T^k(n) bounded above by C for an infinite number > of numbers n. > > And even if you could prove that there are no nontrivial cylces bounded > above by C, this type of argument would be on par with someone arguing > that the Pythagorean Theorem is wrong by claiming that if one were to > construct a huge right triangle with sides (a,b,c) in terms with an > area on the order of hundreds of square miles where a^2+b^2=c^2, then > this would disprove the Pythagorean Theorem and therefore the > Pythagorean Theorem is wrong because the proof of the Pythagorean > Theorem never explicitly considered this possibility. > I don't understand your point here. The Pythagorean Theorem is a > theorem (look up what that means). Collatz is still a conjecture. > It is indeed an open question whether Collatz becomes false if > the numbers are big enough. That CANNOT happen with Pythagoras > regardless of what was and was not considered. > > And saying that my proof is invalid because I never considered the > possibility of using math techniques such as mathematical induction and > proof by contradiction is also on par with this type of argument that > the Pythagorean Theorem is wrong. > This analogy makes no sense. Pythagoras has been PROVED. > Collatz has not. You have to show that a proof via induction or > contradiction CANNOT exist. You have offered no such evidence. > You have shown only that a proof by YOUR methods cannot exist > and you have simply ASSUMED that proofs by other methods cannot > exist. Assuming your premise is true is NOT proof that it is true. > My proof that Collatz is unprovable proves that all methods including > proof by induction and proof by contradiction are useless for trying to > prove the Collatz Conjecture. No it doesn't. You've only shown that methods involving an infinite number of steps are unprovable (duh). That does not apple to proofs by induction and contradiction. > The people who think that my proof is > invalid because it doesn't explicitly mention these types of techniques > by name are talking just like the people who lived a little after > Galois' time and said that his impossibility proof for finding roots of > 5th degree polynomials was invalid because it didn't explicitly > consider every possible formula in terms of elementary arithmetic > operations for the roots of a general 5th degree polynomial, but he > only considered formulas which fit into his limited model of possible > formulas. The only difference between these two proofs is that my > result is simple and obvious ...and wrong. > and can be easily understood with a little > common sense, while Galois' proof requires a sophisticated brilliant > argument which requires a lot of time to understand. > It's so basic that it's amazing that there are still people out there > who don't believe it - in order to know whether the Collatz algorithm > halts, you have to run the algorithm. No, you don't. > There are no shortcuts. Yes, there are. And when MY paper is published, you'll learn what they are. > It's > really just an example of Stephen Wolfram's notion of computational > irreducibility, Except that it doesn't apply in this case. > as he talks about in his book New Kind of Science. It's > just what he predicts would happen in his book when math problems reach > a certain threshold of complexity and start to involve logical > operations as well as equations. Once you get to that threshold, there > is nothing you can do to predict what an algorithm is going to do > unless you run the algorithm itself. False premise, invalid conclusion. > Ever since my proof that Collatz is unprovable has been up on the > internet, anyone who has read it and still thinks it is possible to > prove Collatz might as well try to find a general formula for the roots > of a fifth degree polynomial in terms of elementary arithmetic > operations as well. Serious math journals which receive papers claiming > a proof of Collatz should now begin rejecting these papers as crank > papers without reading them, since such proofs are impossible. Although my paper doesn't prove Collatz, it DOES demonstrate that such proofs ARE possible. By the way, what journal has published your proof? > Craig > > Craig > > --- Christopher Heckman === Subject: Re: Another Reason Why Collatz is Unprovable > I think the context of my argument that Collatz is unprovable is > clear - What I mean by unprovable is unprovable in any reasonable > axiom system. This would exclude inconsistent axiom systems. The > fact that I never specified the axiom system does not take away from > my argument. If Collatz is true, adding it as an axiom does not make the system inconsistent. === Subject: SF: Finally useful theory? I pondered why all those algebraic manipulations just kept giving me equations that relied on my target and realized it was obvious, so I thought about how to get something different, and had a realization. Let S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) where S is the surrogate, and you have to multiply everything out and complete the square, like usual. But if my idea is a good one, you'll end up with this complicated expression with k_1, k_2, k_3 and k_4 multipled times S, where you would then pick that complicated piece to equal your target composite. Then you just pick any S, like S=15, and get squares for x and y using it, and then use the sign ambiguity of the square roots. You see, one way the solution will factor your surrogate S. The other way, changing the signs, it will factor your target, if this idea works. Um, can someone solve out that equation for S to get the complicated thing at the end? It may have special properties that are there just to block this idea!!! If this idea works, it will be a beautiful demonstration of the inherent sign ambiguity in the square root, as what I thought up is to get an expression that is forced to factor more than one number by use of square roots. James Harris === Subject: Re: JSH: Finally useful theory? [added JSH: to subject, spared sci.crypt] [jstevh@msn.com] > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, You managed to do so without realizing that people have _told_ you why, over & over & over again, for more than a year now? That's the first thing that came to mind when you later accused someone else: > Why are you so stupid? I have to explain everything to you in such > detail. Pot, meet the mother of all kettles. > so I thought about how to get something different, and had a realization. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. Huh? You expect x and y to vanish, or what? > Then you just pick any S, like S=15, and get squares for x and y using > it, and then use the sign ambiguity of the square roots. Huh? > You see, one way the solution will factor your surrogate S. > The other way, changing the signs, it will factor your target, if this > idea works. Huh? > Um, can someone solve out that equation for S to get the complicated > thing at the end? I have the software, but can't make much sense of your words. After multiplying out, isolating sqrt(x*y) on one side, and squaring to get rid of the radical: (s - k_3*k_1*x - k_4*k_2*y)^2 = (k_4*k_1 + k_3*k_2)^2*y*x What do you want to do next? For example, multiply it all out again and complete the square wrt x? If so, then we're at (with x only on the LHS, inside a square -- but all the other symbols are in there too): (2*k_3*k_1*s - 2*(k_3)^2*(k_1)^2*x + (k_4)^2*(k_1)^2*y + (k_3)^2*(k_2)^2*y)^2 = (k_4*k_1 + k_3*k_2)^2 * y * (4*k_3*k_1*s + (k_4)^2*(k_1)^2*y - 2*k_4*k_3*k_2*k_1*y + (k_3)^2*(k_2)^2*y) and this is starting to look like a fool's errand. If you then want to expand the mess on the RHS and complete the square wrt y: 2 2 2 2 2 2 2 (2 k_3 k_1 s - 2 k_3 k_1 x + k_4 k_1 y + k_3 k_2 y) = 2 - (-(k_4 k_1 + k_3 k_2) 2 2 2 2 2 (2 k_3 k_1 s + k_4 k_1 y - 2 k_4 k_3 k_2 k_1 y + k_3 k_2 y) 2 2 4 2 3 3 2 4 2 2 2 + 4 k_4 k_3 k_1 s + 8 k_4 k_3 k_2 k_1 s + 4 k_3 k_2 k_1 s ) 2 /(k_4 k_1 - k_3 k_2) Good luck with that :-) > ... === Subject: Re: JSH: Finally useful theory? [jstevh@msn.com] > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, > You managed to do so without realizing that people have _told_ you why, over > & over & over again, for more than a year now? That's the first thing that > came to mind when you later accused someone else: > Why are you so stupid? I have to explain everything to you in such > detail. > Pot, meet the mother of all kettles. > so I thought about how to get something different, and had a realization. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. > Huh? You expect x and y to vanish, or what? > Then you just pick any S, like S=15, and get squares for x and y using > it, and then use the sign ambiguity of the square roots. > Huh? > You see, one way the solution will factor your surrogate S. > The other way, changing the signs, it will factor your target, if this > idea works. > Huh? > Um, can someone solve out that equation for S to get the complicated > thing at the end? > I have the software, but can't make much sense of your words. After > multiplying out, isolating sqrt(x*y) on one side, and squaring to get rid of > the radical: > (s - k_3*k_1*x - k_4*k_2*y)^2 = (k_4*k_1 + k_3*k_2)^2*y*x > What do you want to do next? For example, multiply it all out again and > complete the square wrt x? If so, then we're at (with x only on the LHS, > inside a square -- but all the other symbols are in there too): > (2*k_3*k_1*s - 2*(k_3)^2*(k_1)^2*x + (k_4)^2*(k_1)^2*y + > (k_3)^2*(k_2)^2*y)^2 > (k_4*k_1 + k_3*k_2)^2 * > y * > (4*k_3*k_1*s + (k_4)^2*(k_1)^2*y - 2*k_4*k_3*k_2*k_1*y + > (k_3)^2*(k_2)^2*y) > and this is starting to look like a fool's errand. If you then want to > expand the mess on the RHS and complete the square wrt y: > 2 2 2 2 2 2 2 > (2 k_3 k_1 s - 2 k_3 k_1 x + k_4 k_1 y + k_3 k_2 y) = > 2 > - (-(k_4 k_1 + k_3 k_2) > 2 2 2 2 2 > (2 k_3 k_1 s + k_4 k_1 y - 2 k_4 k_3 k_2 k_1 y + k_3 k_2 y) > 2 2 4 2 3 3 2 4 2 2 2 > + 4 k_4 k_3 k_1 s + 8 k_4 k_3 k_2 k_1 s + 4 k_3 k_2 k_1 s ) > 2 > /(k_4 k_1 - k_3 k_2) > Good luck with that :-) > ... Rather hard to read. Next step is to set that last piece--the piece that doesn't have x or y in it--to T, your target, and see if you can do so without knowing the factorization of T ahead of time. Wait, are some of the k's raised to the 4th power? If so that may be the block I was expecting, how the algebra keeps this idea from being useful for factoring. Oh well, another dead idea. I'm not surprised. James Harris === Subject: Re: JSH: Finally useful theory? [jstevh@msn.com] > ... > Rather hard to read. Hard in a proportional font, easy in a fixed-width font. > Next step is to set that last piece--the piece that doesn't have x or y > in it--to T, your target, and see if you can do so without knowing the > factorization of T ahead of time. The rearranged expression was of the general form: X^2 = (Y^2 + MESS)/D where Y doesn't contain x, and MESS & D don't contain x or y. If you split off the MESS/D part it can factored as (and this is simple enough that I'll switch to a notation that should be readable enough regardless of font): -[2*k_1*k_3*(k_1*k_4 + k_2*k_3)*S / (k_1*k_4 - k_2*k_3)]^2 > Wait, are some of the k's raised to the 4th power? When fully expanded, yes. > ... === Subject: Re: JSH: Finally useful theory? ... > Rather hard to read. > Hard in a proportional font, easy in a fixed-width font. > Next step is to set that last piece--the piece that doesn't have x or y > in it--to T, your target, and see if you can do so without knowing the > factorization of T ahead of time. > The rearranged expression was of the general form: > X^2 = (Y^2 + MESS)/D Yup. Got that part. What you call MESS has S^2 multiplied times an expression with only the k's. That part needs to be set to T, the target to be factored. > where Y doesn't contain x, and MESS & D don't contain x or y. If you split > off the MESS/D part it can factored as (and this is simple enough that I'll > switch to a notation that should be readable enough regardless of font): > -[2*k_1*k_3*(k_1*k_4 + k_2*k_3)*S / (k_1*k_4 - k_2*k_3)]^2 > Wait, are some of the k's raised to the 4th power? > When fully expanded, yes. I don't see that. And why is it S instead of S^2? In any event, you set the part that doesn't have S, to T, your target, and try to find integers k_1, k_2, k_3 and k_4 such that you get that target. So, the MESS turns into TS^2. Got it yet? If you can do that then you will find that T is (I guess to you unless you comprehend the theory) math-a-magically factored by using some value for S, like S=15. Do you understand the theory yet? Do you know why T would be factored if that MESS is set to T? It is so freaking brilliant, and I am increasingly certain the BFC problem is solved. Trouble is, how do you solve the quartics? Can it be done? James Harris === Subject: Re: JSH: Finally useful theory? [Tim Peters] ... >> The rearranged expression was of the general form: >> X^2 = (Y^2 + MESS)/D [jstevh@msn.com] > Yup. Got that part. What you call MESS has S^2 multiplied times an > expression with only the k's. Yes, while D has only k's. > That part needs to be set to T, the target to be factored. >> where Y doesn't contain x, and MESS & D don't contain x or y. If you >> split off the MESS/D part it can factored as (and this is simple enough >> that I'll switch to a notation that should be readable enough regardless >> of font): >> -[2*k_1*k_3*(k_1*k_4 + k_2*k_3)*S / (k_1*k_4 - k_2*k_3)]^2 > Wait, are some of the k's raised to the 4th power? >> When fully expanded, yes. > I don't see that. And why is it S instead of S^2? Look at the factored expression above more carefully? The entire thing is the negation of a square; it's of the form: -Z^2 You get back S^2 if you square the whole thing inside the outer square brackets. You get back fourth powers if you similarly multiply out the factored innards and square the result. Here, I'll expand it as far as is possible: - (4 * (k_1)^4 * (k_3)^2 * (k_4)^2 * S^2)/ ((k_1)^2 * (k_4)^2 - 2 * k_1 * k_2 * k_3 * k_4 + (k_2)^2 * (k_3)^2) - (8 * (k_1)^3 * k_2 * (k_3)^3 * k_4 * S^2)/ ((k_1)^2 * (k_4)^2 - 2 * k_1 * k_2 * k_3 * k_4 + (k_2)^2 * (k_3)^2) - (4 * (k_1)^2 * (k_2)^2 * (k_3)^4 * S^2)/ ((k_1)^2 * (k_4)^2 - 2 * k_1 * k_2 * k_3 * k_4 + (k_2)^2 * (k_3)^2) That's equal to the one-liner factored expression quoted above. So is this: - (2^2 *(k_1)^2*(k_3)^2*(k_1*k_4 + k_2*k_3)^2 * S^2) / (k_1*k_4 - k_2*k_3)^2 I expect you can parse at least one of those > In any event, you set the part that doesn't have S, to T, your target, > and try to find integers k_1, k_2, k_3 and k_4 such that you get that > target. > So, the MESS turns into TS^2. > Got it yet? So you want T = MESS/S^2 = -[2*k_1*k_3*(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3)]^2 > If you can do that then you will find that T is (I guess to you unless > you comprehend the theory) math-a-magically factored by using some > value for S, like S=15. > Do you understand the theory yet? No. > Do you know why T would be factored if that MESS is set to T? If you're restricting the k_i to integers, and can choose them such that 2*k_1*k_3*(k_1*k_4 + k_2*k_3) / (k_1*k_4 - k_2*k_3) is an integer, obviously -- but that can't be made to work unless T itself is the square of an integer (I'm ignoring the minus sign, since I doubt it matters). > It is so freaking brilliant, and I am increasingly certain the BFC > problem is solved. > Trouble is, how do you solve the quartics? As above, the expression factors nicely into the negation of a square. > Can it be done? Because MESS and D are themselves squares, not generally if you're restricting the k_i to be integers. === Subject: Re: SF: Finally useful theory? Originator: gordon@hammy.burditt.org (Gordon Burditt) >I pondered why all those algebraic manipulations just kept giving me >equations that relied on my target and realized it was obvious, so I >thought about how to get something different, and had a realization. You are missing one of the most important issues with Quantum Surrogate Factoring: if you post before you finish the calculation, it will mess them up and it won't work. So try this: 1. Work out the equations. 2. DON'T POST IT. 3. Demonstrate how to factor something simple, like 6 or 32111. Consider using the surrogate value of DON'T POST IT squared. 4. DON'T POST IT. 5. Check that you come out with correct factors by multiplying them together (this you should be able to do with pencil and paper). If the factors come out wrong, go back to 1. 6. DON'T POST IT. 7. Demonstrate how to factor one of the RSA challenge numbers. 8. DON'T POST IT. 9. Check that you come out with correct factors by multiplying them together. If the factors come out wrong, go back to 1. 10. DON'T POST IT. 11. Claim the prize money. 12. DON'T POST IT. 13. Receive the prize money. 14. NOW you can post the stuff from (1), (3), (5), (7), (9), and an image of the check from (13) if you want, and start gloating. Until this step, the liars in sci.math and sci.crypt won't even consider your theory to be worth lying about. Gordon L. Burditt >Let >S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >where S is the surrogate, and you have to multiply everything out and >complete the square, like usual. >But if my idea is a good one, you'll end up with this complicated >expression with k_1, k_2, k_3 and k_4 multipled times S, where you >would then pick that complicated piece to equal your target composite. >Then you just pick any S, like S=15, and get squares for x and y using >it, and then use the sign ambiguity of the square roots. >You see, one way the solution will factor your surrogate S. >The other way, changing the signs, it will factor your target, if this >idea works. >Um, can someone solve out that equation for S to get the complicated >thing at the end? >It may have special properties that are there just to block this >idea!!! >If this idea works, it will be a beautiful demonstration of the >inherent sign ambiguity in the square root, as what I thought up is to >get an expression that is forced to factor more than one number by use >of square roots. >James Harris === Subject: Re: SF: Finally useful theory? <128ov5ecrj2sif2@corp.supernews.comI pondered why all those algebraic manipulations just kept giving me >equations that relied on my target and realized it was obvious, so I >thought about how to get something different, and had a realization. > You are missing one of the most important issues with Quantum Surrogate > Factoring: if you post before you finish the calculation, it will mess > them up and it won't work. So try this: > 1. Work out the equations. > 2. DON'T POST IT. > 3. Demonstrate how to factor something simple, like 6 or 32111. > Consider using the surrogate value of DON'T POST IT squared. > 4. DON'T POST IT. I deleted the rest of your ranting. You clearly don't know how this idea works yet, but at least you did the first piece!!! S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) Here S is NOT the target to be factored but the surrogate. The target comes in later with that expression where you have all the k's which interested people can get some grasp of from your previous post--muddled as the presentation was. If you can set the piece multiplied by S^2 in your post to T, then you incidentally factor T, with some trivial factorization of S, like with S=15. It's a brilliant idea. If it escapes you fine. I get a brilliant factoring idea which is just a little complicated and you are completely out of your element. Why am I not surprised? Matters not. More brilliant minds can look at your beginning analysis in this thread to get some grasp of how it works. Note, his post is on sci.math only I think. Or you can use math software with S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) where you multiply it out, get rid of the radicals, and complete the square twice to see what I mean. Then you just have to deal with quartics, where you have 3 degrees of freedom with the k's, and then program it, optimize it, and go collect your RSA prize money. James Harris === Subject: Re: SF: Finally useful theory? > DON'T POST IT. ^^^ What he said. === Subject: Re: SF: Finally useful theory? | > DON'T POST IT. | | ^^^ What he said. ^^^ What he said === Subject: Re: SF: Finally useful theory? <128ov5ecrj2sif2@corp.supernews.com> <4f%ig.428$6P2.326@fe07.lga> DON'T POST IT. > | ^^^ What he said. > ^^^ What he said Ignore posts that don't interest you. It's a public forum. Freedom of speech goes both ways. If you start limiting some people because it makes sense to your limited intellect, you head down a slippery slope to where just about anyone can be limited, and freedom of speech is out the window. Usenet allows a beautiful way for handling people who don't interest you. You don't read their posts. I don't email television stations bothering them about stupid shows that irritate me. I don't watch them. I don't call up executives at radio stations asking them why they put on stupid shows that annoy me. I don't listen to them. What is wrong with people on Usenet who cannot accept responsibility for their own decision to read someone's posts, so they throw free speech out the window and start ordering people what to say or not say? (Unless of course you're haranguing other people for ordering you what to say. ) I think you're stupid if you do that. I think you're publicly stupid in an area that goes dead if enough people start listening to stupid people like you who want to control what other people say. If you tell other people what to post, when they're not just being evil, like pushing hatred and bigotry or being obnoxious by trying to tell you what to say like what just happened to me, then you are anti-Usenet as that is an attempt to control their FREE SPEECH on a PUBLIC FORUM. IT IS NOT YOUR HOME. YOU MAY BE SITTING AT HOME. BUT USENET IS NOT YOUR FREAKING HOME. IT IS A PUBLIC SPACE. Treat it as such. Do you go to public parks and order people what to do? James Harris === Subject: Re: SF: Finally useful theory? evil, like pushing hatred and bigotry or being obnoxious by trying to > tell you what to say like what just happened to me, then you are > anti-Usenet as that is an attempt to control their FREE SPEECH on a > PUBLIC FORUM. > IT IS NOT YOUR HOME. YOU MAY BE SITTING AT HOME. BUT USENET IS NOT > YOUR FREAKING HOME. > IT IS A PUBLIC SPACE. > Treat it as such. Do you go to public parks and order people what to > do? stay on point, where is the math? Or you just going to bitch all the time? === Subject: Re: SF: Finally useful theory? <128ov5ecrj2sif2@corp.supernews.com> <4f%ig.428$6P2.326@fe07.lga> <448cc20b$0$16347$892e7fe2@authen.yellow.readfreenews.net evil, like pushing hatred and bigotry or being obnoxious by trying to > tell you what to say like what just happened to me, then you are > anti-Usenet as that is an attempt to control their FREE SPEECH on a > PUBLIC FORUM. > IT IS NOT YOUR HOME. YOU MAY BE SITTING AT HOME. BUT USENET IS NOT > YOUR FREAKING HOME. > IT IS A PUBLIC SPACE. > Treat it as such. Do you go to public parks and order people what to > do? > stay on point, where is the math? Or you just going to bitch all the time? Ok. Let S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) where S is the surrogate, and you have to multiply everything out and complete the square, like usual. I looked to an expression with square roots because the square root is ambiguous in that it has two solutions--a positive and a negative one. That way that expression can't equal just a single S, where S this time is the surrogate and not T the target. I'm hoping someone will use some math software and give the solution, which involves multiplying everything out, squaring to get rid of sqrt(xy), and then completing the square twice, which I think will give you this monstrous expression that has some complicated looking thing with k_1, k_2, k_3 and k_4 multiplied times S. You set that complicated thing to your target, unless the math has a way to block you, or forces you to know the factorization of your target ahead of time. If someway, somehow this idea is valid, and you can DO all of that without knowing your target's factorization ahead of time, then something almost magical will happen next, as you pick S, like S = 15, then, for instance, you can have k_1*sqrt(x) + k_2*sqrt(y) = 5 k_3*sqrt(x) + k_4*sqrt(y) = 3 and you find squares for x and y that work. And then you just change signs, and, well, if it works, then you factor your target, incidentally. If it works it's an exceedingly clever idea that actually uses the ambiguity of the square root, to factor, if it works. James Harris === Subject: Re: SF: Finally useful theory? >> If you tell other people what to post, when they're not just being >> evil, like pushing hatred and bigotry or being obnoxious by trying to >> tell you what to say like what just happened to me, then you are >> anti-Usenet as that is an attempt to control their FREE SPEECH on a >> PUBLIC FORUM. >> IT IS NOT YOUR HOME. YOU MAY BE SITTING AT HOME. BUT USENET IS NOT >> YOUR FREAKING HOME. >> IT IS A PUBLIC SPACE. >> Treat it as such. Do you go to public parks and order people what to >> do? >> stay on point, where is the math? Or you just going to bitch all the >> time? > Ok. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > I looked to an expression with square roots because the square root is > ambiguous in that it has two solutions--a positive and a negative one. > That way that expression can't equal just a single S, where S this time > is the surrogate and not T the target. > I'm hoping someone will use some math software and give the solution, > which involves multiplying everything out, squaring to get rid of > sqrt(xy), and then completing the square twice, which I think will give > you this monstrous expression that has some complicated looking thing > with k_1, k_2, k_3 and k_4 multiplied times S. > You set that complicated thing to your target, unless the math has a > way to block you, or forces you to know the factorization of your > target ahead of time. > If someway, somehow this idea is valid, and you can DO all of that > without knowing your target's factorization ahead of time, then > something almost magical will happen next, as you pick S, like S = 15, > then, for instance, you can have > k_1*sqrt(x) + k_2*sqrt(y) = 5 > k_3*sqrt(x) + k_4*sqrt(y) = 3 > and you find squares for x and y that work. try x = 4 and y=9 then k_1*2 + k_2*3 = 5 k_3*2 + k_4*9 = 3 I have 4 unknowns and two equns............... now what ? === Subject: Re: SF: Finally useful theory? <128ov5ecrj2sif2@corp.supernews.com> <4f%ig.428$6P2.326@fe07.lga> <448cc20b$0$16347$892e7fe2@authen.yellow.readfreenews.net> <448cf2a1$0$15296$892e7fe2@authen.yellow.readfreenews.net> If you tell other people what to post, when they're not just being >> evil, like pushing hatred and bigotry or being obnoxious by trying to >> tell you what to say like what just happened to me, then you are >> anti-Usenet as that is an attempt to control their FREE SPEECH on a >> PUBLIC FORUM. >> IT IS NOT YOUR HOME. YOU MAY BE SITTING AT HOME. BUT USENET IS NOT >> YOUR FREAKING HOME. >> IT IS A PUBLIC SPACE. >> Treat it as such. Do you go to public parks and order people what to >> do? >> stay on point, where is the math? Or you just going to bitch all the >> time? > Ok. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > I looked to an expression with square roots because the square root is > ambiguous in that it has two solutions--a positive and a negative one. > That way that expression can't equal just a single S, where S this time > is the surrogate and not T the target. > I'm hoping someone will use some math software and give the solution, > which involves multiplying everything out, squaring to get rid of > sqrt(xy), and then completing the square twice, which I think will give > you this monstrous expression that has some complicated looking thing > with k_1, k_2, k_3 and k_4 multiplied times S. > You set that complicated thing to your target, unless the math has a > way to block you, or forces you to know the factorization of your > target ahead of time. > If someway, somehow this idea is valid, and you can DO all of that > without knowing your target's factorization ahead of time, then > something almost magical will happen next, as you pick S, like S = 15, > then, for instance, you can have > k_1*sqrt(x) + k_2*sqrt(y) = 5 > k_3*sqrt(x) + k_4*sqrt(y) = 3 > and you find squares for x and y that work. > try x = 4 and y=9 > then k_1*2 + k_2*3 = 5 > k_3*2 + k_4*9 = 3 > I have 4 unknowns and two equns............... now what ? You have to have the k's. A Tim Peters made a post in this thread to only sci.math where he stepped through the first piece. If you multiply out S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) isolate the sqrt(xy) to one side and square both sides to clear out radicals, and then complete the square twice, you get this rather complicated expression, where you can define the k's by your target. Trouble is, the defining expression has k's raised to the 4th power, so you have a quartic to solve. The mathematics here is kind of complicated. Complicated enough that is far beyond the complexity that I tend to like, so it may escape you. It is best to use math software to get the equations as S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) explodes in complexity as you step through what I described. It's kind of neat actually. Such a simple expression to start which just kind of blows up as you begin the analysis. Kind of wild actually. James Harris === Subject: Re: SF: Finally useful theory? Usenet is not a free space. It is comprised of newsgroups about specific subjects, such as cryptology. Your posts are not about cryptology. Your posts are (at best) your diarrhetic, mathematically invalid, brainstorming. Up until now all you have shown us that z=z and (supposedly) from this stock markets around the world will crash from this knowledge. I tremble in fear. /sarcasm === Subject: Re: SF: Finally useful theory? > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, so I > thought about how to get something different, and had a realization. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. > Then you just pick any S, like S=15, and get squares for x and y using > it, and then use the sign ambiguity of the square roots. > You see, one way the solution will factor your surrogate S. > The other way, changing the signs, it will factor your target, if this > idea works. If you change them both, then yes, it works: (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) = (-k_1*sqrt(x) - k_2*sqrt(y))*(- k_3*sqrt(x) - k_4*sqrt(y)), since you're multiplying each factor by -1, and (-x)*(-y) = (-1)*(-1)*x*y = x * y. But if you change just one, you will change the value of S (unless you change +0 to -0, or vice versa). > Um, can someone solve out that equation for S to get the complicated > thing at the end? S = (k_1 k_3 x + k_2 k_4 y) + (k_2 k_3 + k_1 k_4) sqrt(xy) The modified S is: (k_1 k_3 x + k_2 k_4 y) - (k_2 k_3 + k_1 k_4) sqrt(xy). > It may have special properties that are there just to block this idea!!! You mean mathematics is censoring itself??? That's a keeper!!! > If this idea works, it will be a beautiful demonstration of the > inherent sign ambiguity in the square root, as what I thought up is to > get an expression that is forced to factor more than one number by use > of square roots. That's a big IF, though. --- Christopher Heckman === Subject: Re: SF: Finally useful theory? > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, so I > thought about how to get something different, and had a realization. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. > Then you just pick any S, like S=15, and get squares for x and y using > it, and then use the sign ambiguity of the square roots. > You see, one way the solution will factor your surrogate S. > The other way, changing the signs, it will factor your target, if this > idea works. > If you change them both, then yes, it works: > (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > = (-k_1*sqrt(x) - k_2*sqrt(y))*(- k_3*sqrt(x) - k_4*sqrt(y)), > since you're multiplying each factor by -1, and > (-x)*(-y) = (-1)*(-1)*x*y = x * y. > But if you change just one, you will change the value of S (unless you > change +0 to -0, or vice versa). > Um, can someone solve out that equation for S to get the complicated > thing at the end? > S = (k_1 k_3 x + k_2 k_4 y) + (k_2 k_3 + k_1 k_4) sqrt(xy) > The modified S is: > (k_1 k_3 x + k_2 k_4 y) - (k_2 k_3 + k_1 k_4) sqrt(xy). > It may have special properties that are there just to block this idea!!! > You mean mathematics is censoring itself??? That's a keeper!!! No you dimwit. You have to solve the equations out for x and y using TWO completions of the square. If my guess is right you'll get this rather complicated expression at the very end, which has k_1, k_2, k_3, k_4 and S. If you can get that expression to have your target as a factor, then necessarily any factorization of S, will factor your target in a weird bit of clever mathematics. My feeling though is that the algebra is up to this trick and will make k_1, k_2, k_3, and k_4 dependent on the factorization of your target, so that this approach is useless. It should be obvious though if you solve everything out, which is a tedious calculation so I haven't done it. Why are you so stupid? I have to explain everything to you in such detail. And you think you are a mathematician? James Harris === Subject: Re: SF: Finally useful theory? > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, so I > thought about how to get something different, and had a realization. > > Let > > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. > > Then you just pick any S, like S=15, and get squares for x and y using > it, and then use the sign ambiguity of the square roots. > > You see, one way the solution will factor your surrogate S. > > The other way, changing the signs, it will factor your target, if this > idea works. > If you change them both, then yes, it works: > (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > = (-k_1*sqrt(x) - k_2*sqrt(y))*(- k_3*sqrt(x) - k_4*sqrt(y)), > since you're multiplying each factor by -1, and > (-x)*(-y) = (-1)*(-1)*x*y = x * y. > But if you change just one, you will change the value of S (unless you > change +0 to -0, or vice versa). > Um, can someone solve out that equation for S to get the complicated > thing at the end? > S = (k_1 k_3 x + k_2 k_4 y) + (k_2 k_3 + k_1 k_4) sqrt(xy) > The modified S is: > (k_1 k_3 x + k_2 k_4 y) - (k_2 k_3 + k_1 k_4) sqrt(xy). > It may have special properties that are there just to block this idea!!! > You mean mathematics is censoring itself??? That's a keeper!!! > No you dimwit. Gratuitious insult! Now, you can't complain when I say: Your bulb went out _long_ ago, buddy. > You have to solve the equations out for x and y using > TWO completions of the square. You can either solve for x in terms of k_1, k_2, k_3, k_4, S, AND y, or solve for y in terms of k_1, k_2, k_3, k_4, S, and x. You only have one equation, so there's no hope in getting x and y both in terms of the k_i's and S. > If my guess is right you'll get this rather complicated expression at > the very end, which has k_1, k_2, k_3, k_4 and S. > If you can get that expression to have your target as a factor, then > necessarily any factorization of S, will factor your target in a weird > bit of clever mathematics. > My feeling though is that the algebra is up to this trick and will make > k_1, k_2, k_3, and k_4 dependent on the factorization of your target, > so that this approach is useless. > It should be obvious though if you solve everything out, which is a > tedious calculation so I haven't done it. > Why are you so stupid? I have to explain everything to you in such > detail. You have to explain everything to _everybody_. That suggests that it is not my problem, but rather your explanations which are causing the confusion. I don't have any problems like this with my students, even the college algebra classes. --- Christopher Heckman === Subject: Re: SF: Finally useful theory? > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, so I > thought about how to get something different, and had a realization. > > Let > > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. > > Then you just pick any S, like S=15, and get squares for x and y using > it, and then use the sign ambiguity of the square roots. > > You see, one way the solution will factor your surrogate S. > > The other way, changing the signs, it will factor your target, if this > idea works. > > If you change them both, then yes, it works: > > (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > = (-k_1*sqrt(x) - k_2*sqrt(y))*(- k_3*sqrt(x) - k_4*sqrt(y)), > > since you're multiplying each factor by -1, and > (-x)*(-y) = (-1)*(-1)*x*y = x * y. > > But if you change just one, you will change the value of S (unless you > change +0 to -0, or vice versa). > > Um, can someone solve out that equation for S to get the complicated > thing at the end? > > S = (k_1 k_3 x + k_2 k_4 y) + (k_2 k_3 + k_1 k_4) sqrt(xy) > > The modified S is: > > (k_1 k_3 x + k_2 k_4 y) - (k_2 k_3 + k_1 k_4) sqrt(xy). > > It may have special properties that are there just to block this idea!!! > > You mean mathematics is censoring itself??? That's a keeper!!! > No you dimwit. > Gratuitious insult! Yup. I admit it. You, um, not so smart person. > Now, you can't complain when I say: Your bulb went out _long_ ago, > buddy. Well, coming from you... The idea here is clearly escaping you. I have a case where S is NOT the target. So I have all my key variables defined by a couple of equations where the target is nowhere to be seen, which solves my BFC problem. (BFC stands for Big Freaking Circle) > You have to solve the equations out for x and y using > TWO completions of the square. > You can either solve for x in terms of k_1, k_2, k_3, k_4, S, AND y, or > solve for y in terms of k_1, k_2, k_3, k_4, S, and x. You only have one > equation, so there's no hope in getting x and y both in terms of the > k_i's and S. If you can't even get to the point where you just have the k's. Hey, Tim Peters made a post on sci.math where he stepped through the first piece with some math software so you can freaking SEE what I mean, which may save me explaining and explaining to your, um, to you how it works. Look at Peters's post and see how it's done. James Harris === Subject: Re: SF: Finally useful theory? > I have to explain everything to you in such detail. Perhaps you forgot 8th grade algebra, when you had to do those ever-so-annoying things called proofs...? === Subject: Re: SF: Finally useful theory? > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, so I > thought about how to get something different, and had a realization. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. In case you don't get it. You pick S, like S=15. You find squares that will work for x and y. Then you shift the signs. That means you get two additional solutions. That is the perfect part of the idea. > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. That is the unknown part of the idea. Can you pick k_1, k_2, k_3 and k_4 such that multiplied times your surrogate S is your target T? I don't know. But if you can, then that's it. The factoring problem is solved in a damn clever way. James Harris === Subject: Re: SF: Finally useful theory? > I pondered why all those algebraic manipulations just kept giving me > equations that relied on my target and realized it was obvious, so I > thought about how to get something different, and had a realization. > Let > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, and you have to multiply everything out and > complete the square, like usual. > In case you don't get it. You pick S, like S=15. No, you pick T first. Let's choose T = 21. > You find squares that will work for x and y. This sentence has no meaning. For instance, there's no way to write it in Java. > Then you shift the signs. > That means you get two additional solutions. But now you might get trivial factors. You might find out that 21 = 1 * 21 = 21 * 1, which does not help you at all. > That is the perfect part of the idea. > But if my idea is a good one, you'll end up with this complicated > expression with k_1, k_2, k_3 and k_4 multipled times S, where you > would then pick that complicated piece to equal your target composite. > That is the unknown part of the idea. > Can you pick k_1, k_2, k_3 and k_4 such that multiplied times your > surrogate S is your target T? That sentence does not parse. You need a noun between that and multiplied. > I don't know. > But if you can, then that's it. The factoring problem is solved in a > damn clever way. You've been saying this for years. You've posted some procedure which is too vague to encode. You've called people liars when they try to figure out what you're saying and find a counterexample. You've found out on your own that they were right. You've been wrong every single time. You should work out your ideas before you post them. You should read through your posts before posting them. --- Christopher Heckman === Subject: Re: SF: Finally useful theory? OpenPGP: id=A0E28D18 > That sentence does not parse. May I point you to Chomsky's result that there cannot even be parser for simplified English? > You need a noun between that and multiplied. === Subject: Re: SF: Finally useful theory? >>That sentence does not parse. > May I point you to Chomsky's result that there cannot even be parser for > simplified English? >>You need a noun between that and multiplied. This not a sentence. === Subject: A Formula for T^k(n): Was: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at [...] > Once again, let U be the modified Collatz function > U(n) = n/2, if n is even, > U(n) = (n-1)/2, if n is odd. > [...] > In this case, it is implicit in the argument that there is a one-to-one > correspondence between all of the possible formulas for T^k and all > possible values of (n,T(n),...,T^(k-1)(n)) mod 2 that the possible > formulas for T^k cannot be reduced any further for any k. > [whereas they can for U^k(n)]. But T^k(n) can also be written as a single formula, in terms of n and k. This suggests that you need to define formula in your paper. First of all, recall that: T(n) = n/2, if n is even. T(n) = (3n+1)/2, if n is odd. PROPOSITION 1. For any integer n, T(n) = n/2 + (n/2 - floor(n/2)) * 2 * (2n+1)/2. Proof. If n is even, then n/2 = floor(n/2), so T(n) is n/2 plus zero times some number, which is n/2. If n is odd, then n/2 - floor(n/2) = 1/2, so T(n) = n/2 + (1/2) *2 * (2n+1)/2 = n/2 + (2n+1)/2 = (3n+1)/2. QED. PROPOSITION 2. For any integer n, and any nonnegative integer k, there is a finite formula for T^k(n), using the arithmetic operations and the floor function. Proof. Induction on k. If k=0, then this formula is n. If we have a formula for T^k(n), then T^(k+1)(n) = T(T^k(n)) = [T^k(n)]/2 + ([T^k(n)]/2 - floor([T^k(n)]/2)) * 2 * (2[T^k(n)] + 1)/2, by Proposition 1. QED. PROPOSITION 3. There is a formula F such that F(n,k) = T^k(n), for all integers n, and all nonnegative integers k. Proof. One such formula is sum (0^abs(k-i) * F_i(n,i), i=0..infinity), where F_i(n,i) is the finite formula for T^i(n) guaranteed by Proposition 2. (0^x = 0 if x > 0, and 0^0 = 1. If you don't like this convention, you can use 1 - ceiling(arctan(abs(k - i))) instead of 0^abs(k-i).) The infinite series above converges for all n, k --- in fact, at most one term is nonzero, even if k is not an integer. That single term will be when i=k, and will simplify to F_k(n,k), which is T^k(n). QED. As to whether this formula can be reduced, well, that's another question ... --- Christopher Heckman === Subject: JSH: Pure math, factoring idea Well if this latest idea of mine actually can be made to work, it will be so damn ironic that it will be hard to comprehend that irony. Over the years I've gotten into stupid arguments with math people on this forum about the inherent ambiguity of the sqrt() as it HAS to give two answers, no matter what people define. Well, looking at S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) I used that reality in a very, very clever way, if it works!!! If you multiply everything out, eliminate the radical by squaring, and then complete the square twice, you should get this rather complicated expression, which is what is factored by solutions for x and y. Being a clever person you pick S as your surrogate, so you know squares for x and y that will work!!! But then you get a bonus! Because there are TWO solutions to the square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless there's some block as I haven't looked at the complicated expression) you may end up factoring a target composite. If this idea works, you might be able to factor an RSA sized number using S=15. And that would also be rather ironic. I guess I really should have math software that can do that calculation. I'm not even going to try it just yet, hoping someone else will, as I'm so likely to make errors doing it by hand. It is a bit of a horrendous calculation to do by hand. James Harris === Subject: Re: Pure math, factoring idea > Well if this latest idea of mine actually can be made to work, it will > be so damn ironic that it will be hard to comprehend that irony. > Over the years I've gotten into stupid arguments with math people on > this forum about the inherent ambiguity of the sqrt() as it HAS to give > two answers, no matter what people define. > Well, looking at > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > I used that reality in a very, very clever way, if it works!!! > If you multiply everything out, eliminate the radical by squaring, and > then complete the square twice, you should get this rather complicated > expression, which is what is factored by solutions for x and y. > Being a clever person you pick S as your surrogate, so you know squares > for x and y that will work!!! > But then you get a bonus! Because there are TWO solutions to the > square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless > there's some block as I haven't looked at the complicated expression) > you may end up factoring a target composite. > If this idea works, you might be able to factor an RSA sized number > using S=15. > And that would also be rather ironic. > I guess I really should have math software that can do that > calculation. I'm not even going to try it just yet, hoping someone > else will, as I'm so likely to make errors doing it by hand. > It is a bit of a horrendous calculation to do by hand. trivial only took me a page. === Subject: Re: Pure math, factoring idea <448c1a88$0$16347$892e7fe2@authen.yellow.readfreenews.net Well if this latest idea of mine actually can be made to work, it will > be so damn ironic that it will be hard to comprehend that irony. > Over the years I've gotten into stupid arguments with math people on > this forum about the inherent ambiguity of the sqrt() as it HAS to give > two answers, no matter what people define. > Well, looking at > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > I used that reality in a very, very clever way, if it works!!! > If you multiply everything out, eliminate the radical by squaring, and > then complete the square twice, you should get this rather complicated > expression, which is what is factored by solutions for x and y. > Being a clever person you pick S as your surrogate, so you know squares > for x and y that will work!!! > But then you get a bonus! Because there are TWO solutions to the > square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless > there's some block as I haven't looked at the complicated expression) > you may end up factoring a target composite. > If this idea works, you might be able to factor an RSA sized number > using S=15. > And that would also be rather ironic. > I guess I really should have math software that can do that > calculation. I'm not even going to try it just yet, hoping someone > else will, as I'm so likely to make errors doing it by hand. > It is a bit of a horrendous calculation to do by hand. > trivial > only took me a page. Still I make dumb mistakes with such calculations. I take it then you found that k_1, k_2, k_3 and k_4 are related to the factorization of your target? That's what I expect to be true. It is kind of neat, pure math kind of thing though. A way to express two factorizations using just one. I guess none of you want to play along any more which is why no one posted the answer. And you really shouldn't do such a calculation by hand when math software is so good at it, and---perfect. James Harris === Subject: Re: Pure math, factoring idea <448c1a88$0$16347$892e7fe2@authen.yellow.readfreenews.net > Well if this latest idea of mine actually can be made to work, it will > be so damn ironic that it will be hard to comprehend that irony. > > Over the years I've gotten into stupid arguments with math people on > this forum about the inherent ambiguity of the sqrt() as it HAS to give > two answers, no matter what people define. > > Well, looking at > > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > > I used that reality in a very, very clever way, if it works!!! > > If you multiply everything out, eliminate the radical by squaring, and > then complete the square twice, you should get this rather complicated > expression, which is what is factored by solutions for x and y. > > Being a clever person you pick S as your surrogate, so you know squares > for x and y that will work!!! > > But then you get a bonus! Because there are TWO solutions to the > square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless > there's some block as I haven't looked at the complicated expression) > you may end up factoring a target composite. > > If this idea works, you might be able to factor an RSA sized number > using S=15. > > And that would also be rather ironic. > > I guess I really should have math software that can do that > calculation. I'm not even going to try it just yet, hoping someone > else will, as I'm so likely to make errors doing it by hand. > > It is a bit of a horrendous calculation to do by hand. > trivial > only took me a page. > Still I make dumb mistakes with such calculations. > I take it then you found that k_1, k_2, k_3 and k_4 are related to the > factorization of your target? > That's what I expect to be true. > It is kind of neat, pure math kind of thing though. A way to express > two factorizations using just one. > I guess none of you want to play along any more which is why no one > posted the answer. LIAR! I multiplied it out, by hand. > And you really shouldn't do such a calculation by hand when math > software is so good at it, and---perfect. Actually, the time to do it by hand is much less than it takes to start up Maple or Mathematica, type it in, and get the answer. Any college algebra student should be able to do it; it's just a use of the FOIL technique. I suppose next JSH will say we shouldn't do 3*4 by hand. --- Christopher Heckman === Subject: Re: Pure math, factoring idea >> Well if this latest idea of mine actually can be made to work, it >> will >> be so damn ironic that it will be hard to comprehend that irony. >> Over the years I've gotten into stupid arguments with math people on >> this forum about the inherent ambiguity of the sqrt() as it HAS to >> give >> two answers, no matter what people define. >> Well, looking at >> S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >> I used that reality in a very, very clever way, if it works!!! >> If you multiply everything out, eliminate the radical by squaring, >> and >> then complete the square twice, you should get this rather >> complicated >> expression, which is what is factored by solutions for x and y. >> Being a clever person you pick S as your surrogate, so you know >> squares >> for x and y that will work!!! >> But then you get a bonus! Because there are TWO solutions to the >> square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless >> there's some block as I haven't looked at the complicated expression) >> you may end up factoring a target composite. >> If this idea works, you might be able to factor an RSA sized number >> using S=15. >> And that would also be rather ironic. >> I guess I really should have math software that can do that >> calculation. I'm not even going to try it just yet, hoping someone >> else will, as I'm so likely to make errors doing it by hand. >> It is a bit of a horrendous calculation to do by hand. >> trivial >> only took me a page. >> Still I make dumb mistakes with such calculations. >> I take it then you found that k_1, k_2, k_3 and k_4 are related to the >> factorization of your target? >> That's what I expect to be true. >> It is kind of neat, pure math kind of thing though. A way to express >> two factorizations using just one. >> I guess none of you want to play along any more which is why no one >> posted the answer. > LIAR! I multiplied it out, by hand. >> And you really shouldn't do such a calculation by hand when math >> software is so good at it, and---perfect. > Actually, the time to do it by hand is much less than it takes to start > up Maple or Mathematica, type it in, and get the answer. > Any college algebra student should be able to do it; it's just a use of > the FOIL technique. > I suppose next JSH will say we shouldn't do 3*4 by hand. JSH is only a troll. nothing more. === Subject: Re: JSH: Pure math, factoring idea (Quoting it in its entirety, to preserve it.) > Well if this latest idea of mine actually can be made to work, it will > be so damn ironic that it will be hard to comprehend that irony. > Over the years I've gotten into stupid arguments with math people on > this forum about the inherent ambiguity of the sqrt() as it HAS to give > two answers, no matter what people define. > Well, looking at > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > I used that reality in a very, very clever way, if it works!!! > If you multiply everything out, eliminate the radical by squaring, and > then complete the square twice, you should get this rather complicated > expression, which is what is factored by solutions for x and y. > Being a clever person you pick S as your surrogate, so you know squares > for x and y that will work!!! > But then you get a bonus! Because there are TWO solutions to the > square root, if you pick k_1, k_2, k_3 and k_4 correctly (unless > there's some block as I haven't looked at the complicated expression) > you may end up factoring a target composite. Those are some big IF's. Even bigger than: ******** ********** * * * * *** *** * ******* * * * * * * * * * * * ******* * * * * * * * ******* * * * * * * * * * * * * *** *** * * * * * * ******** **** > If this idea works, you might be able to factor an RSA sized number > using S=15. Try using it to factor T=21 with S=15 before you get your hopes up. But you can't, because you haven't said how you're assuming T factors, or how T is related to S. > And that would also be rather ironic. > I guess I really should have math software that can do that > calculation. I'm not even going to try it just yet, hoping someone > else will, as I'm so likely to make errors doing it by hand. > It is a bit of a horrendous calculation to do by hand. > James Harris === Subject: Re: JSH: SF: Finally, surrogate factoring [snipped to the essentials] [Rick Decker guesses at what JSH meant, a seemingly impossible task] [Tim Peters] >> Are you psychic or what? [Rick Decker] > Yes, and I knew you'd ask that. Damn. I'd have to be an idiot not to see that one coming, and I didn't. Worse: ... >> Yes, you are psychic! > I knew you'd say that. I have to admit it. Even after getting caught above, I didn't see this one coming either. But you knew that too, so on to the remaining tech stuff: >> I had no idea how he came up with the thing containing 21g_1, and >> never would have guessed he was just pulling it out of his butt :-) > Surely you're not surprised. No -- I may be a non-psychic idiot, but I still learn from experience. ... > Sadly, I predict you're right again. I'll let that stand alone, since it's right pretty much regardless of context === Subject: Re: JSH: SF: Finally, surrogate factoring [jstevh@msn.com] >> You're both lying. [Proginoskes] > Well, suppose for the moment that that is true. Since mathematics > cannot be hidden, the lie _must_ be visible in the post above. Which > step, precisely, is the lie in, James? He probably doesn't remember anymore <0.5 wink>. I'm 90% sure he didn't bother to read the post beyond Rick's initial summary of what was established several posts earlier: [Rick Decker] So completing the square w.r.t y first yields (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T Completing the square w.r.t. z first yields (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T Rick and I used software to derive those (independently, using Mathematica and Macsyma respectively). The algebra is too tedious for James to do correctly by hand, and he's apparently too lazy to even bother plugging in specific values to _check_ whether they're lies (a check he could easily have performed against his own flawed derivation to learn at once that it was wrong). Anyway, he hates that these derivations are both of the form: A^2 = B^2 + i*T because he started with T=g_1*g_2, and in both cases above carrying on with _easy_ algebra shows that in the derived (A+B)*(A-B)= i*T, A+B and A-B are just trivial respellings of j*g_1 and k*g_2 for some fixed integers j and k with j*k=i. IOW, if you believe Rick or me, this method is essentially useless unless you know T's factorization before you start. Well, it's useless then for a reason James _understands_; there are other reasons for why a method of this form is almost certainly useless regardless, but try explaining those to our James Since he can neither do the math himself nor be bothered to check the result when it's handed to him, but _can_ understand that the result shows his method is useless, his conclusion is that we're lying about that result. He can't show an error in the derivation because (a) there isn't one; and, (b) his only real reason for believing it's a lie is that it shoots down his factoring wet dream du jour. Assume you're right. Then I'm wrong. That contradicts that I'm right, therefore you're wrong. More, since I'm so obviously right, you must be lying. QED >> I am going to warn you. >> Neither of you will live to see 2007 because of this lie becaue there >> will be angry people who will kill you, as there is so much money at >> stake. >> Billions will be lost. >> It's not like you can reverse it now either. >> You killed yourselves. >> It might have seemed like a small lie to both of you, but your names >> will live in infamy, while you will not live at all. I apologize for quoting that part yet again, but it so astonishes me each time I read it I just can't help wanting to share my delight anew :-) === Subject: Something a little more lighthearted! http://pi.ytmnd.com/ :) -lynX === Subject: Re: Something a little more lighthearted! > http://pi.ytmnd.com/ > :) How come they don't sing the whole song? > -lynX === Subject: Re: Something a little more lighthearted! > http://pi.ytmnd.com/ > :) > How come they don't sing the whole song? -lynX Ahaha, they should! I was kindof upset too. :) -lynX === Subject: basic question on affine algebraic varieties Suppose k is any algebraically closed field. For f a polynomial in k[x_1,x_2,...x_n], V(f) denotes the set of zeros of f. Is it true that V(f) = V(g) ==> f = cg for some c in k? dan === Subject: Re: basic question on affine algebraic varieties : Suppose k is any algebraically closed field. For f a polynomial in : k[x_1,x_2,...x_n], V(f) denotes the set of zeros of f. Is it true that : V(f) = V(g) ==> f = cg for some c in k? What are the zeroes of f^2? Ted === Subject: Re: basic question on affine algebraic varieties : k[x_1,x_2,...x_n], V(f) denotes the set of zeros of f. Is it true that > : V(f) = V(g) ==> f = cg for some c in k? > What are the zeroes of f^2? > Ted You're right, of course, but let me rephrase the question; a) if f and g are irreducible and V(f) = V(g) then must f = cg for some c in k? b) if f and g are square-free and V(f) is contained in V(g) then does it follow that g divides f? dan === Subject: Re: basic question on affine algebraic varieties :> : Suppose k is any algebraically closed field. For f a polynomial in :> : k[x_1,x_2,...x_n], V(f) denotes the set of zeros of f. Is it true that :> : V(f) = V(g) ==> f = cg for some c in k? :> What are the zeroes of f^2? :> Ted : You're right, of course, but let me rephrase the question; : a) if f and g are irreducible and V(f) = V(g) then must f = cg for some : c in k? : b) if f and g are square-free and V(f) is contained in V(g) then does : it follow that g divides f? Yes. (a) follows from (b), and (b) follows from the Nullstellensatz (you only need g to be squarefree). Ted === Subject: Re: basic question on affine algebraic varieties : Suppose k is any algebraically closed field. For f a polynomial in > :> : k[x_1,x_2,...x_n], V(f) denotes the set of zeros of f. Is it true that > :> : V(f) = V(g) ==> f = cg for some c in k? > : :> What are the zeroes of f^2? > : :> Ted > : You're right, of course, but let me rephrase the question; > : a) if f and g are irreducible and V(f) = V(g) then must f = cg for some > : c in k? > : b) if f and g are square-free and V(f) is contained in V(g) then does > : it follow that g divides f? > Yes. (a) follows from (b), and (b) follows from the Nullstellensatz (you > only need g to be squarefree). > Ted dan === Subject: Re: basic question on affine algebraic varieties :> : a) if f and g are irreducible and V(f) = V(g) then must f = cg for some :> : c in k? :> : b) if f and g are square-free and V(f) is contained in V(g) then does :> : it follow that g divides f? :> Yes. (a) follows from (b), and (b) follows from the Nullstellensatz (you :> only need g to be squarefree). :> Ted Actually, in (b), the conclusion should be that f divides g. Ted === Subject: oarnge glow due to U oxide Mail-To-News-Contact: abuse@dizum.com 0 Apple panting === Subject: Need some help with digits... Caveat: I am not a mathematician. I had a couple semester of math on the college level. (They never appeared to deal with anything interesting ;) I'm trying to understand the concept of a digit as it is being used in math. The context is cantor's diagonal argument. Let me lay it out as I understand it, maybe there's already something wrong with my understanding. Suppose there were some interesting list of real numbers that purports to contain all rael numbers. You could write that list down and then construct a new number by taking the n'th digit of the n'th number and increasing it by one or setting it to zero if it is nine. This newly constructed number would have a first digit which is different from the first digit of the first number on the list, its second digit would differ from the second digit of the second number, its third digit would differ from the third digit of the third number and so forth. And therefore the new number can't be on this list as it is different from each of the numbers by at least one digit. (Someone correct me if I have mangled that). Now I have always taken digits to be something ... dunno ... profane. Something to do with writing down numbers which in themselves exist in any representation we can imagine (including none at all). In particular, it is utterly obvious that the same number can be expressed in a multitude of ways ( 1/4 vs 0.25 vs. a quarter etc). My confusion stems from the fact that it is not clear to me that two numbers are necessary unequal, just because they don't use the same digits. For example there's the classical case of 0.999...=1.000... where I'm expressing the same number using two different decimal expressions. Pushing this to an extreme: imagine Cantor's list was really very boring and consisted only of the numbers {1.000..., 1.000..., 1.000...} then I could claim that the number 0.999... is the diagonal number of this list and proclaim to have proven that 0.999... is in fact NOT equal to 1.000... because its first digit is different from that of the first number on the list, its second digit is different from the second digit of the second number etc. Obviously there's something wrong here, but I'm kinda stuck with this. So I figured I'd throw this out here and see if someone can whack me over the head with the (undoubtedly obvious) mistake I'm making here somewhere... codially Y.T. -- http://unwantedinsights.blogspot.com/ Remove YourClothes before you email me. === Subject: Re: Need some help with digits... > Caveat: I am not a mathematician. I had a couple > semester of math on the college level. (They never > appeared to deal with anything interesting ;) Caveat: I am a mathematician in training. I've had a couple of semesters of college level mathematics as well (most of them never went too in-depth with anything of interest ;) > I'm trying to understand the concept of a digit as > it is being used in math. The context is cantor's > diagonal argument. Let me lay it out as I understand > it, maybe there's already something wrong with my > understanding. Since Denis & Virgil (and probably many others to come) have already adressed your understanding, I just wanted to throw my two cents into the pot. Commonly, the digits of a number n in base b are the integers d_i < b that satisfy n = d_k * b^k + ... + d_1 * b^1 + d_0 * b^0 ..and the number n in base b is occasionally written as [d_k ... d_1 d_0]_b. For example, if n = [120]_3 (i.e. the number n in base 3 is represented as 120) then n = 1 * 3^2 + 2 * 3^1 + 0 * 3^0 = 9 + 6 + 0 = 15 ..and we say that the digits of the integer 15 in base 3 are d_2 = 1, d_1 = 2, and d_0 = 0. It's important to note that every real number has a unique decimal expansion. > Suppose there were some interesting list of real > numbers that purports to contain all rael numbers. You > could write that list down and then construct a new > number by taking the n'th digit of the n'th number and > increasing it by one or setting it to zero if it is > nine. As the others have pointed out, we can use many other digit substitutions instead of this one. Another common way is to suppose you have a complete set of all the real numbers in base 2 and proceed via the diagonal argument using 1 |-> 0 and 0 |-> 1. > This newly constructed number would have a first > digit which is different from the first digit of the > first number on the list, its second digit would differ > from the second digit of the second number, its > third digit would differ from the third digit of the > third number and so forth. And therefore the new number > can't be on this list as it is different from each of > the numbers by at least one digit. > (Someone correct me if I have mangled that). > Now I have always taken digits to be something ... > dunno ... profane. Something to do with writing down > numbers which in themselves exist in any representation > we can imagine (including none at all). In particular, > it is utterly obvious that the same number can be > expressed in a multitude of ways ( 1/4 vs 0.25 vs. > a quarter etc). Profane huh o_O interesting; I've always thought of the digits of a number almost as coordinates. > My confusion stems from the fact that it is not clear > to me that two numbers are necessary unequal, just > because they don't use the same digits. For example > there's the classical case of 0.999...=1.000... > where I'm expressing the same number using two > different decimal expressions. Uhhh ohhhh, this makes me recall the billions of threads on this subject. > Pushing this to an extreme: imagine Cantor's list was > really very boring and consisted only of the numbers > {1.000..., 1.000..., 1.000...} then I could claim that > the number 0.999... is the diagonal number of this > list and proclaim to have proven that 0.999... is in > fact NOT equal to 1.000... because its first digit is > different from that of the first number on the list, > its second digit is different from the second digit of > the second number etc. > Obviously there's something wrong here, but I'm kinda > stuck with this. So I figured I'd throw this out here > and see if someone can whack me over the head with the > (undoubtedly obvious) mistake I'm making here > somewhere... Again, as others have pointed out this apparent paradox can be avoided by using other substitution rules. Another problem is that you're writing 1.000..., 1.000..., etc. to represent different numbers, but then you magically arrive at 0.999... and conclude that this was in your list. The trouble here is that 1.000...(all 0's) = 0.999... (all 9's), and so the 1.000... numbers in your set must all have different expansions; that is to say they must have an expansion something like 1.000...00d00... where d =/= 0. Thus, if you're assuming a complete set, you'll eventually arrive at a real number 1.000... which does not contain a zero on it's diagonal (for otherwise this number would not be in your set; a contradiction). Therefore, by your substitution rule, you'd arrive at a number, say 0.999...995999...., which is *not* the same as 0.999...(9's forever) = 1.000...(0's forever). Either way all of this is irrelevant because Cantor's diagonal argument assumes a *complete* set, not the very boring one you proposed. Kyle Czarnecki === Subject: Re: Need some help with digits... > Caveat: I am not a mathematician. I had a couple semester of math on > the college level. (They never appeared to deal with anything > interesting ;) > I'm trying to understand the concept of a digit as it is being used > in math. The context is cantor's diagonal argument. Let me lay it out > as I understand it, maybe there's already something wrong with my > understanding. > Suppose there were some interesting list of real numbers that purports > to contain all rael numbers. You could write that list down and then > construct a new number by taking the n'th digit of the n'th number and > increasing it by one or setting it to zero if it is nine. Actually, one wants a rule which will not ever replace a digit with either a 0 or a 9, as these might lead to numbers with more than one decimal representation. An acceptable rule would be to replace any digit except 1 by 1 and replace 1 by 2. > This newly constructed number would have a first digit which is > different from the first digit of the first number on the list, its > second > digit would differ from the second digit of the second number, its > third digit would differ from the third digit of the third number and > so > forth. And therefore the new number can't be on this list as it is > different from each of the numbers by at least one digit. > (Someone correct me if I have mangled that). > Now I have always taken digits to be something ... dunno > ... profane. Something to do with writing down numbers which in > themselves exist in any representation we can imagine (including none > at all). In particular, it is utterly obvious that the same number can > be expressed in a multitude of ways ( 1/4 vs 0.25 vs. a quarter etc). > My confusion stems from the fact that it is not clear to me that two > numbers are necessary unequal, just because they don't use the same > digits. For example there's the classical case of 0.999...=1.000... > where I'm expressing the same number using two different decimal > expressions. As noted above, one can assure that this situation does not arise. Most numbers have unique decimal representations. Those that have more than one representation gain that ambiguity by either having all but finitely many digits being 0 or all but finitely many being 9. Using a rule like that above, one can assure that the newly created number contains neither of these digits at all. > Pushing this to an extreme: imagine Cantor's list was really very > boring and consisted only of the numbers {1.000..., 1.000..., 1.000...} > then I could claim that the number 0.999... is the diagonal number > of this list and proclaim to have proven that 0.999... is in fact NOT > equal to 1.000... because its first digit is different from that of > the first number on the list, its second digit is different from the > second digit of the second number etc. Not if you use a rule which prohibits the use of either 0 or 9 in the diagonal number. > Obviously there's something wrong here, but I'm kinda stuck with > this. So I figured I'd throw this out here and see if someone can > whack me over the head with the (undoubtedly obvious) mistake I'm > making here somewhere... The problem is real if one is careless about how one constructs the new number, but is easily solved with a little forethought. === Subject: Re: Need some help with digits... [...] > My confusion stems from the fact that it is not clear to me that two > numbers are necessary unequal, just because they don't use the same > digits. For example there's the classical case of 0.999...=1.000... > where I'm expressing the same number using two different decimal > expressions. > As noted above, one can assure that this situation does not arise. > Most numbers have unique decimal representations. > Those that have more than one representation gain that ambiguity by > either having all but finitely many digits being 0 or all but finitely > many being 9. Hum -- OK, this appears to be the crux of the matter. What you write there makes sense to me, but there have been a lot of things in my life that have made sense to me which later turned out to be nonsense, so let me be a little paranoid about your statements there :) In essence, let me propose that the last two sentences of yours that I'm quoting above constitute a theorem. The question then becomes: where can I find more information about this - ideally including a proof for it? When I use the term base, I then to think in terms of an orthogonal (or at least linearly independent) base in the context of a vector space (or some other metric space? It's been years) and then it is obvious that two different representations refer to different elements pretty much by definition of linearly independent. But {1, 10, 100, 1000..} are not linearly independent as I can write 10 as different linear combinations of 1 and 100 (9*1+0.01*100 or 5*1+0.05*100 or...). Now obviously 0.05 isn't a valid digit in a base-10 system and that's probably where my confusion stems from. (In particular the classic 0.999...=1.000... looks like there's something fishy with this linear independence here somewhere). I'll have to ponder that for a while... cordially Y.T. -- Remove YourClothes before you email me. === Subject: Re: Need some help with digits... > [...] > My confusion stems from the fact that it is not clear to me that two > numbers are necessary unequal, just because they don't use the same > digits. For example there's the classical case of 0.999...=1.000... > where I'm expressing the same number using two different decimal > expressions. > As noted above, one can assure that this situation does not arise. > Most numbers have unique decimal representations. > Those that have more than one representation gain that ambiguity by > either having all but finitely many digits being 0 or all but finitely > many being 9. > Hum -- OK, this appears to be the crux of the matter. > What you write there makes sense to me, but there have been a lot of > things in my life that have made sense to me which later turned out > to be nonsense, so let me be a little paranoid about your statements > there :) > In essence, let me propose that the last two sentences of yours that > I'm quoting above constitute a theorem. The question then becomes: > where can I find more information about this - ideally including a > proof for it? > When I use the term base, I then to think in terms of an orthogonal > (or at least linearly independent) base in the context of a vector > space (or some other metric space? It's been years) and then it is > obvious that two different representations refer to different elements > pretty much by definition of linearly independent. In vector spaces one speaks of a basis of a space which is quite a different thing from the base of a positional number system. Any conflation of these two meanings is bound to cause confusion. === Subject: Re: Need some help with digits... ytyourclothes@p.zapto.org a .8ecrit : > Caveat: I am not a mathematician. I had a couple semester of math on > the college level. (They never appeared to deal with anything > interesting ;) > I'm trying to understand the concept of a digit as it is being used > in math. The context is cantor's diagonal argument. Let me lay it out > as I understand it, maybe there's already something wrong with my > understanding. > Suppose there were some interesting list of real numbers that purports > to contain all rael numbers. You could write that list down and then > construct a new number by taking the n'th digit of the n'th number and > increasing it by one or setting it to zero if it is nine. Actually, this is not the exact usual recipe: to prevent the problem you see below, just replace the digit by a 3 if it is not a 3, and replace a 3 by ta 5 > This newly constructed number would have a first digit which is > different from the first digit of the first number on the list, its > second > digit would differ from the second digit of the second number, its > third digit would differ from the third digit of the third number and > so > forth. And therefore the new number can't be on this list as it is > different from each of the numbers by at least one digit. > (Someone correct me if I have mangled that). Ok, apart for this problem with the nines > Now I have always taken digits to be something ... dunno > ... profane. Something to do with writing down numbers which in > themselves exist in any representation we can imagine (including none > at all). In fact, this is the reason some people prefer other proofs of Cantor result, which dont use any special representation. But they tend to be more abstract, while the diagonal argument explicitly exhibit a number not on the list In particular, it is utterly obvious that the same number can > be expressed in a multitude of ways ( 1/4 vs 0.25 vs. a quarter etc). > My confusion stems from the fact that it is not clear to me that two > numbers are necessary unequal, just because they don't use the same > digits. For example there's the classical case of 0.999...=1.000... > where I'm expressing the same number using two different decimal > expressions. Yes. But this is the only possible ambiguity ( with the similar 0.1249999...=0.125) So dont use 9 and 0 at all in the recipe above, and you are safe. > Pushing this to an extreme: imagine Cantor's list was really very > boring and consisted only of the numbers {1.000..., 1.000..., 1.000...} > then I could claim that the number 0.999... is the diagonal number > of this list and proclaim to have proven that 0.999... is in fact NOT > equal to 1.000... because its first digit is different from that of > the first number on the list, its second digit is different from the > second digit of the second number etc. > Obviously there's something wrong here, but I'm kinda stuck with > this. So I figured I'd throw this out here and see if someone can > whack me over the head with the (undoubtedly obvious) mistake I'm > making here somewhere... > codially > Y.T. > -- > http://unwantedinsights.blogspot.com/ > Remove YourClothes before you email me. === Subject: functional equation+derivable Let f be differentiable at x = 1 function which satisfies f(x^3) + f(x^2) + f(x) = x^3 + x^2 + x. Find all such functions f. Any ideas are welcome. === Subject: Re: functional equation+derivable > Let f be differentiable at x = 1 function which satisfies > f(x^3) + f(x^2) + f(x) = x^3 + x^2 + x. > Find all such functions f. > Any ideas are welcome. f(x) = x f(0) = 0; f(1) = 1 f(1) - 2f(-1) = -1; f(-1) = -1 g(x) = f(x) - x; f(x) = g(x) + x Find all g with g(x^3) + g(x^2) + g(x) = 0 g(0) = 0; g(1) = 0; g(-1) = 0 2g(-1) + g(1) = 0 g(x^3) + g(x^2) + g(x) = 0 3x^2 g'(x^3) + 2x.g'(x^2) + g'(x) = 0 6x.g'(x^3) + 9x^4 g(x^3) + 2.g'(x^2) + 4x^2 g(x^2) + g(x) = 0 ... g^(n) (0) = 0, for all n in N. g^(n) (1) = 0, for all n in N. === Subject: Re: functional equation+derivable > Let f be differentiable at x = 1 function which > satisfies > f(x^3) + f(x^2) + f(x) = x^3 + x^2 + x. > Find all such functions f. > Any ideas are welcome. > f(x) = x > f(0) = 0; f(1) = 1 > f(1) - 2f(-1) = -1; f(-1) = -1 > g(x) = f(x) - x; f(x) = g(x) + x > Find all g with > g(x^3) + g(x^2) + g(x) = 0 > g(0) = 0; g(1) = 0; g(-1) = 0 > 2g(-1) + g(1) = 0 > g(x^3) + g(x^2) + g(x) = 0 > 3x^2 g'(x^3) + 2x.g'(x^2) + g'(x) = 0 > 6x.g'(x^3) + 9x^4 g(x^3) + 2.g'(x^2) + 4x^2 g(x^2) > + g(x) = 0 Could you exlain thi point. Why can we differentate g twice ? It was only supposed that g is 1-time differentiable at the x = 1. > g^(n) (0) = 0, for all n in N. > g^(n) (1) = 0, for all n in N. === Subject: Re: functional equation+derivable <31703424.1150019263703.JavaMail.jakarta@nitrogen.mathforum.org > Let f be differentiable at x = 1 function which satisfies > > f(x^3) + f(x^2) + f(x) = x^3 + x^2 + x. for all x in R ? > Find all such functions f. > Any ideas are welcome. > f(x) = x > f(0) = 0; f(1) = 1 > f(1) - 2f(-1) = -1; f(-1) = -1 > g(x) = f(x) - x; f(x) = g(x) + x > Find all g with > g(x^3) + g(x^2) + g(x) = 0 That's a simpler, yet equivalent function search, is it not? > g(0) = 0; g(1) = 0; g(-1) = 0 > 2g(-1) + g(1) = 0 > g(x^3) + g(x^2) + g(x) = 0 > 3x^2 g'(x^3) + 2x.g'(x^2) + g'(x) = 0 > 6x.g'(x^3) + 9x^4 g(x^3) + 2.g'(x^2) + 4x^2 g(x^2) + g(x) = 0 > Could you explain thi point. Why can we differentiate g twice ? It was > only supposed that g is 1-time differentiable at the x = 1. It shows for other than trivial g, there's no smooth answers as indicated (via induction) in the next two lines: > g^(n) (0) = 0, for all n in N. > g^(n) (1) = 0, for all n in N. === Subject: Re: functional equation+derivable > Let f be differentiable at x = 1 function which >> satisfies > f(x^3) + f(x^2) + f(x) = x^3 + x^2 + x. > Find all such functions f. > Any ideas are welcome. >> f(x) = x >> f(0) = 0; f(1) = 1 >> f(1) - 2f(-1) = -1; f(-1) = -1 >> g(x) = f(x) - x; f(x) = g(x) + x >> Find all g with >> g(x^3) + g(x^2) + g(x) = 0 >> g(0) = 0; g(1) = 0; g(-1) = 0 >> 2g(-1) + g(1) = 0 >> g(x^3) + g(x^2) + g(x) = 0 >> 3x^2 g'(x^3) + 2x.g'(x^2) + g'(x) = 0 >> 6x.g'(x^3) + 9x^4 g(x^3) + 2.g'(x^2) + 4x^2 g(x^2) >> + g(x) = 0 > Could you exlain thi point. Why can we differentate g twice ? It was only supposed that g is 1-time differentiable at the x = 1. You are right in asking this as it is the crucial point for your original question. Here is a sketch how to see that g is twice differentiable around x=1: Represent g(1+h)=h r(h) (*) where r is a function bounded around 0. Now it suffices to show that r(h)/h is bounded around 0, too, by plugging (*) into 0=g(x^3)+g(x^2)+g(x), sorting by powers of h, dividing by h^2 and having a very close look the relation resulting from that. I didn't check the details, but I think it works. Best wishes, J. === Subject: Over $250 Free Poker, Casino Money - No deposits required Hi all math fans, check out http://www.freepokeraccounts.com for over $250 dollars in gambling money. You know the odds, or I guess you can figure them out.. so why risk using your own cash?! :) You sign up for the free poker cash at say titan, party poker, CDPoker, etc.. Then you can use the money on any of the games they offer... blackjack, slots, craps, roulette, video poker, hold'em, Omaha etc.. Hit up these deals before they are gone! This is a great deal and not some scam or a trick, so check out the website today for more information. (Notice: you will never see another post about this on this newsgroup, I just feel some of you might like to know about these deals) === Subject: Re: Over $250 Free Poker, Casino Money - No deposits required > Hi all math fans, check out http://www.freepokeraccounts.com for over > $250 dollars in gambling money. You know the odds, or I guess you can > figure them out.. so why risk using your own cash?! :) > You sign up for the free poker cash at say titan, party poker, CDPoker, > etc.. 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James WHO? === Subject: Simpson Integration - a specific problem Normally Simpson is used for integrating between an interval a to b, by using the end points of the interval (a and b) as nodes. An advantage of this procedure is, that by doubling the number of nodes between two iteration loops the already calculated nodes can be reused, so that only about half of the nodes have to be calculated new. Due to a singularity at one of the end points of the interval, I try to apply the Simpson integration slightly different. I split the integration interval a to b in in equidistant subintervals and use the middle of these intervals as nodes. If, in this procedure, the number of nodes would be doubled one would loos the possibility of reusing already calculated nodes. This disadvantage can be overcame by trebling the number of nodes from one iteration loop to the next, i.e. in an iteration loop one third of the nodes can be used from the previous loop and two thirds have to be calculated new. This adjusted procedure calculates the integral, but, contrary of what I would have expected, with a very poor efficiency. Comparing the required number of iteration loops for a given Rel. Error = ABS( 1 - Result_new_Loop / Result_old_Loop ) between the adjusted Simpson procedure and a basic numerical integration (sums of Fi*dF, doubling nodes without reusing already calculated ones) shows the following differences: Adjusted Simpson Procedure: Rel. Error Result Number of nodes <1E-3 2062.2593 78732 <1E-4 2063.7296 4960116 <1E-5 2063.8892 389014812 Basic iteration procedure Rel. Error Result Number of nodes <1E-5 2063.9027 3072 (approx. * 2) <1E-8 2063.9091 294912 (approx. * 2) Questions: I can somehow understand that the adjusted Simpson procedure requires more iteration loops since the relative error is calculated with twice the number of new nodes compared to the number of old nodes. But why is the result not more accurate with the much larger number of nodes? Or, are the Simpson factors for the sums (1, 4 and 2) not the same for this adjusted version? Is there a possibility to adjust the Simpson integration so that it becomes more efficient than the basic iteration procedure? Marcel === Subject: Re: Simpson Integration - a specific problem >Normally Simpson is used for integrating between an interval a to b, by >using the end points of the interval (a and b) as nodes. An advantage >of this procedure is, that by doubling the number of nodes between two >iteration loops the already calculated nodes can be reused, so that >only about half of the nodes have to be calculated new. >Due to a singularity at one of the end points of the interval, I try to >apply the Simpson integration slightly different. I split the >integration interval a to b in in equidistant subintervals and use the >middle of these intervals as nodes. If, in this procedure, the number >of nodes would be doubled one would loos the possibility of reusing >already calculated nodes. This disadvantage can be overcame by trebling >the number of nodes from one iteration loop to the next, i.e. in an >iteration loop one third of the nodes can be used from the previous >loop and two thirds have to be calculated new. This adjusted procedure >calculates the integral, but, contrary of what I would have expected, >with a very poor efficiency. I'm not sure what formula you're using for your modified Simpson's procedure. Have you tried it on simple integrals with no singularities? Does it work there, giving similar behaviour (error ~ const/n^4) to the usual Simpson's Rule? In general, the success of Simpson's Rule depends on the integrand being fairly smooth, with an error bound depending on the fourth derivative. A singularity at an end point is likely to defeat most numerical methods unless you're careful. Can you compute the singular part and subtract it off before integrating? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Simpson Integration - a specific problem On 11 Jun 2006 02:35:00 -0700, MET The correct place for this is sci.math.num-analysis. >Normally Simpson is used for integrating between an interval a to b, by >using the end points of the interval (a and b) as nodes. An advantage >of this procedure is, that by doubling the number of nodes between two >iteration loops the already calculated nodes can be reused, so that >only about half of the nodes have to be calculated new. >Due to a singularity at one of the end points of the interval, I try to >apply the Simpson integration slightly different. I split the >integration interval a to b in in equidistant subintervals and use the >middle of these intervals as nodes. If, in this procedure, the number >of nodes would be doubled one would loos the possibility of reusing >already calculated nodes. This disadvantage can be overcame by trebling >the number of nodes from one iteration loop to the next, i.e. in an >iteration loop one third of the nodes can be used from the previous >loop and two thirds have to be calculated new. This adjusted procedure >calculates the integral, but, contrary of what I would have expected, >with a very poor efficiency. >Comparing the required number of iteration loops for a given >Rel. Error = ABS( 1 - Result_new_Loop / Result_old_Loop ) >between the adjusted Simpson procedure and a basic numerical >integration (sums of Fi*dF, doubling nodes without reusing already >calculated ones) shows the following differences: >Adjusted Simpson Procedure: >Rel. Error Result Number of nodes ><1E-3 2062.2593 78732 ><1E-4 2063.7296 4960116 ><1E-5 2063.8892 389014812 >Basic iteration procedure >Rel. Error Result Number of nodes ><1E-5 2063.9027 3072 (approx. * 2) ><1E-8 2063.9091 294912 (approx. * 2) >Questions: >I can somehow understand that the adjusted Simpson procedure requires >more iteration loops since the relative error is calculated with twice >the number of new nodes compared to the number of old nodes. But >why is the result not more accurate with the much larger number of >nodes? Or, are the Simpson factors for the sums (1, 4 and 2) not the >same for this adjusted version? Is there a possibility to adjust the >Simpson integration so that it becomes more efficient than the basic >iteration procedure? >Marcel === Subject: Re: Simpson Integration - a specific problem You might get something out of reading Numerical Recipes ch. 4.1 at http://www.library.cornell.edu/nr/cbookcpdf.html . Martin -- Quidquid latine scriptum sit, altum viditur. === Subject: Re: Simpson Integration - a specific problem apply the Simpson integration slightly different. I split the >integration interval a to b in in equidistant subintervals and use the >middle of these intervals as nodes. Please post the integration rule as you are using it, such as some code or pseudocode. Dave === Subject: Re: Simpson Integration - a specific problem apply the Simpson integration slightly different. I split the >integration interval a to b in in equidistant subintervals and use the >middle of these intervals as nodes. > Please post the integration rule as you are using it, such as some code > or pseudocode. > Dave I try to do it . The code is written in Pascal/Delphi. There are two ways of adding comments in this language; one for a comment over several lines which starts with (* and ends with *) and an other one to add a comment in a line: all after // is comment. Note also that the number of subintervalls are increased by a factor of 3; this is also the reason for the value of 3 in the denominator when calculating the relative error. I hope it's correctly copied and understandable. Marcel // Do integration over dR //----------------------- sumF1 := 0.0; sumF2 := 0.0; sumF4 := 0.0; sumF := 99999.99; nCount:=1 repeat // S t a r t: I n t e g r a t i o n - L o o p nCount := nCount * 3 ; old_sumF := sumF; // Move already calculated sums sumFtmp:=sumF1+sumF2; sumF1:=0.0; sumF2:=sumF4; sumF4:=sumFtmp; dR := (Rend-Rstart)/nCount; (* Here: calculation of values ..j which correspond here to the values at the lower limit of the integration interval, i.e. the start point of the integration. *) for iCount:=1 to nCount do begin // i C o u n t (* Here: values ..j become values ..i which are the values at the lower limit of the subinterval *) (* Here: calculation of values ..j which are the values at the upper limit of the subinterval, i.e the values at i+dr *) if ( // first time no existing data: all iCount (nCount=3) or // data exists if (iCount+1) is a multiple of 3 (((iCount+1) mod 3)<>0) ) then begin // C a l c u l a t e (* Calculate the function at the middle of the subinterval: Note that the integrand contains dR=Rj-Ri in the denominator and after the integration also in the nominator (see last line of the code) *) tanZ:=Tan((Zi+Zj)/2); mu:=(mui+muj)/2; dF := tanZ / mu * (muj-mui) / (Rj-Ri); if Odd(iCount) then begin if ( (iCount=1) or (iCount=nCount) ) then sumF1 := sumF1 + dF else sumF2 := sumF2 + dF; end else sumF4 := sumF4 + dF; end; // C a l c u l a t e end; // i C o u n t sumF := - (sumF1 + 2*sumF2 + 4*sumF4); until // E n d : I n t e g r a t i o n - L o o p (ABS(1.0-sumF/old_sumF/3) I should add that at the start of the integration interval z=pi/4 and therefore tan(z) is infinite. Further I should mention that the problem exists also if z is not equal +/- pi/4 in the integration interval, but to a lesser extend. Summing rectangles would still be more efficient. === Subject: Re: Simpson Integration - a specific problem >I should add that at the start of the integration interval z=pi/4 and >therefore tan(z) is infinite. >Further I should mention that the problem exists also if z is not equal >+/- pi/4 in the integration interval, but to a lesser extend. Summing >rectangles would still be more efficient. I thought tan(pi/4)=1? what you want to do is to simulate an open Newton Cotes formula using the closed formula on a shifted grid. this is no good idea. the equivalent open compagnon of Simpson is this one: int_{a to b} f(x)dx = (1/4)*(b-a)/2*(2*f(a+(b-a)/6)+3*f((a+b)/2)+2*f(b-(b-a)/6)) +((b-a)/2)^5*(7/1620)*f^{(4)}(xi) and you can use this one for an integrable singularity, although much better rules exist for this task. here you can reuse all the old function values on a subinterval if you divide a too large subinterval into three equal parts, (but this needs 6 new function values) but should be finally be better than the midpoint rule. the weights must be redefined of course since the relative position of the nodes change their role form inner to outer node. you can use simsons rule if on one side of the interval the integrand is finite: you simply begin with a small subinterval on the left say, applying simpsons rule and refining if the precision is not met for this part. then proceed to the right with the estimation of the new interval size (like this is done for ode's) but using not b but (b-current stepsize)/2 as the new upper bound. integrating towards the singularity will automatically reduce the step, such that you can approach the singularity with arbitrary precision. in quadpack http://www.netlib.org/quadpack singular integrals are always reduced to singular integrals over ]0,1[ by a substitution and then treated by adaptive Gauss/Gauss-Kronrod quadrature (error estimate is the difference between the Gauss and the Gauss-Kornrod value) with refinement where necessary. these nodes are also always inside the interval . hth peter === Subject: Re: Simpson Integration - a specific problem > On 11 Jun 2006 02:35:00 -0700, MET > Normally Simpson is used for integrating between an interval a to b, by >> using the end points of the interval (a and b) as nodes. An advantage >> of this procedure is, that by doubling the number of nodes between two >> iteration loops the already calculated nodes can be reused, so that >> only about half of the nodes have to be calculated new. >> Due to a singularity at one of the end points of the interval, I try to >> apply the Simpson integration slightly different. I split the >> integration interval a to b in in equidistant subintervals and use the >> middle of these intervals as nodes. If, in this procedure, the number >> of nodes would be doubled one would loos the possibility of reusing >> already calculated nodes. This disadvantage can be overcame by trebling >> the number of nodes from one iteration loop to the next, i.e. in an >> iteration loop one third of the nodes can be used from the previous >> loop and two thirds have to be calculated new. This adjusted procedure >> calculates the integral, but, contrary of what I would have expected, >> with a very poor efficiency. >> Comparing the required number of iteration loops for a given >> Rel. Error = ABS( 1 - Result_new_Loop / Result_old_Loop ) >> between the adjusted Simpson procedure and a basic numerical >> integration (sums of Fi*dF, doubling nodes without reusing already >> calculated ones) shows the following differences: >> Adjusted Simpson Procedure: >> Rel. Error Result Number of nodes >> <1E-3 2062.2593 78732 >> <1E-4 2063.7296 4960116 >> <1E-5 2063.8892 389014812 >> Basic iteration procedure >> Rel. Error Result Number of nodes >> <1E-5 2063.9027 3072 (approx. * 2) >> <1E-8 2063.9091 294912 (approx. * 2) >> Questions: >> I can somehow understand that the adjusted Simpson procedure requires >> more iteration loops since the relative error is calculated with twice >> the number of new nodes compared to the number of old nodes. But >> why is the result not more accurate with the much larger number of >> nodes? Or, are the Simpson factors for the sums (1, 4 and 2) not the >> same for this adjusted version? Is there a possibility to adjust the >> Simpson integration so that it becomes more efficient than the basic >> iteration procedure? >> Marcel I strongly recommend a book on numerical analysis, such as Ralston or maybe Acton, Numerical Methods that Work. Also the sections on numerical quadrature in Abramowitz & Stegun. Or you might look at my notes on numerical quadrature at http://galileo.phys.virginia.edu/classes/551.jvn.fall01/551Notes.htm -- Julian V. Noble Professor Emeritus of Physics University of Virginia === Subject: Re: Simpson Integration - a specific problem this specific application I actually can live with just summing rectangles. The question was more related to why the Simpson procedure can be so much inferior to the basic integration procedure. Are there known application limits where the integration by Simpson is by 5 to 6 order of magnitudes less efficient than just summing rectangles? Marcel > On 11 Jun 2006 02:35:00 -0700, MET > Normally Simpson is used for integrating between an interval a to b, by >> using the end points of the interval (a and b) as nodes. An advantage >> of this procedure is, that by doubling the number of nodes between two >> iteration loops the already calculated nodes can be reused, so that >> only about half of the nodes have to be calculated new. >> Due to a singularity at one of the end points of the interval, I try to >> apply the Simpson integration slightly different. I split the >> integration interval a to b in in equidistant subintervals and use the >> middle of these intervals as nodes. If, in this procedure, the number >> of nodes would be doubled one would loos the possibility of reusing >> already calculated nodes. This disadvantage can be overcame by trebling >> the number of nodes from one iteration loop to the next, i.e. in an >> iteration loop one third of the nodes can be used from the previous >> loop and two thirds have to be calculated new. This adjusted procedure >> calculates the integral, but, contrary of what I would have expected, >> with a very poor efficiency. >> Comparing the required number of iteration loops for a given >> Rel. Error = ABS( 1 - Result_new_Loop / Result_old_Loop ) >> between the adjusted Simpson procedure and a basic numerical >> integration (sums of Fi*dF, doubling nodes without reusing already >> calculated ones) shows the following differences: >> Adjusted Simpson Procedure: >> Rel. Error Result Number of nodes >> <1E-3 2062.2593 78732 >> <1E-4 2063.7296 4960116 >> <1E-5 2063.8892 389014812 >> Basic iteration procedure >> Rel. Error Result Number of nodes >> <1E-5 2063.9027 3072 (approx. * 2) >> <1E-8 2063.9091 294912 (approx. * 2) >> Questions: >> I can somehow understand that the adjusted Simpson procedure requires >> more iteration loops since the relative error is calculated with twice >> the number of new nodes compared to the number of old nodes. But >> why is the result not more accurate with the much larger number of >> nodes? Or, are the Simpson factors for the sums (1, 4 and 2) not the >> same for this adjusted version? Is there a possibility to adjust the >> Simpson integration so that it becomes more efficient than the basic >> iteration procedure? >> Marcel > I strongly recommend a book on numerical analysis, such as Ralston > or maybe Acton, Numerical Methods that Work. Also the sections on > numerical quadrature in Abramowitz & Stegun. > Or you might look at my notes on numerical quadrature at > http://galileo.phys.virginia.edu/classes/551.jvn.fall01/551Notes.htm > -- > Julian V. Noble > Professor Emeritus of Physics > University of Virginia === Subject: Re: Simpson Integration - a specific problem > this specific application I actually can live with just summing > rectangles. The question was more related to why the Simpson procedure > can be so much inferior to the basic integration procedure. Are there > known application limits where the integration by Simpson is by 5 to 6 > order of magnitudes less efficient than just summing rectangles? > Marcel > On 11 Jun 2006 02:35:00 -0700, MET > The correct place for this is sci.math.num-analysis. >> Normally Simpson is used for integrating between an interval a to b, by >> using the end points of the interval (a and b) as nodes. An advantage >> of this procedure is, that by doubling the number of nodes between two >> iteration loops the already calculated nodes can be reused, so that >> only about half of the nodes have to be calculated new. >> Due to a singularity at one of the end points of the interval, I try to >> apply the Simpson integration slightly different. I split the >> integration interval a to b in in equidistant subintervals and use the >> middle of these intervals as nodes. If, in this procedure, the number >> of nodes would be doubled one would loos the possibility of reusing >> already calculated nodes. This disadvantage can be overcame by trebling >> the number of nodes from one iteration loop to the next, i.e. in an >> iteration loop one third of the nodes can be used from the previous >> loop and two thirds have to be calculated new. This adjusted procedure >> calculates the integral, but, contrary of what I would have expected, >> with a very poor efficiency. >> Comparing the required number of iteration loops for a given >> Rel. Error = ABS( 1 - Result_new_Loop / Result_old_Loop ) >> between the adjusted Simpson procedure and a basic numerical >> integration (sums of Fi*dF, doubling nodes without reusing already >> calculated ones) shows the following differences: >> Adjusted Simpson Procedure: >> Rel. Error Result Number of nodes >> <1E-3 2062.2593 78732 >> <1E-4 2063.7296 4960116 >> <1E-5 2063.8892 389014812 >> Basic iteration procedure >> Rel. Error Result Number of nodes >> <1E-5 2063.9027 3072 (approx. * 2) >> <1E-8 2063.9091 294912 (approx. * 2) >> Questions: >> I can somehow understand that the adjusted Simpson procedure requires >> more iteration loops since the relative error is calculated with twice >> the number of new nodes compared to the number of old nodes. But >> why is the result not more accurate with the much larger number of >> nodes? Or, are the Simpson factors for the sums (1, 4 and 2) not the >> same for this adjusted version? Is there a possibility to adjust the >> Simpson integration so that it becomes more efficient than the basic >> iteration procedure? >> Marcel >> I strongly recommend a book on numerical analysis, such as Ralston >> or maybe Acton, Numerical Methods that Work. Also the sections on >> numerical quadrature in Abramowitz & Stegun. >> Or you might look at my notes on numerical quadrature at >> http://galileo.phys.virginia.edu/classes/551.jvn.fall01/551Notes.htm >> -- >> Julian V. Noble >> Professor Emeritus of Physics >> University of Virginia When you integrate to a singularity a higher-order formula can be less efficient than a lower-order one. The answer is to factor out the singularity and integrate a less-singular function. For example, sqrt{x} near x=0 is an integrable singularity. So if you can write int _0^1 {f(x)} = int _0^1 {sqrt{x} g(x)} where g(0) = constant, you can rewrite as int _0^1 {sqrt{x} ( g(x) - g(0) )} + 2g(0)/3 and the new integral is better behaved (vanishes faster) at x=0. Alternatively you could use Gauss-Chebyshev near the singularity. See Abramowitz & Stegun. -- Julian V. Noble Professor Emeritus of Physics University of Virginia === Subject: Provinf bounded variation I wish to prove that a function is of bounded variation. The function i'll regard is f(x) = cos(kx) and i have the following suggestion. f in BV([a,b]) if it can be written as a sum of two functions; one non-decreasing and one non-increasing. Here i get unsecure because even if i could piece-up the cosine in monotonous segments, i'd like to use the following set-up. On the segments where cosine increases: +: cos(kx) -: 0 while where cosine decreases: +: -cos(kx) -: 2cos(kx) Is it right to regard the problem this way? Is there a more straightforward approach? (Only yes or no, please. It's sort of a homework.) -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Provinf bounded variation > I wish to prove that a function is of bounded variation. > The function i'll regard is f(x) = cos(kx) and i have the > following suggestion. > f in BV([a,b]) if it can be written as a sum of two > functions; one non-decreasing and one non-increasing. > Here i get unsecure because even if i could piece-up > the cosine in monotonous segments, i'd like to use the > following set-up. > On the segments where cosine increases: > +: cos(kx) -: 0 > while where cosine decreases: > +: -cos(kx) -: 2cos(kx) > Is it right to regard the problem this way? > Is there a more straightforward approach? I think it's easier to use the fact that every C^1 function is BV on bounded intervals; the proof is immediatefrom the mean value theorem. === Subject: Re: Provinf bounded variation >> I wish to prove that a function is of bounded variation. >> The function i'll regard is f(x) = cos(kx) and i have the >> following suggestion. >> f in BV([a,b]) if it can be written as a sum of two >> functions; one non-decreasing and one non-increasing. >> On the segments where cosine increases: >> +: cos(kx) -: 0 >> while where cosine decreases: >> +: -cos(kx) -: 2cos(kx) >> Is it right to regard the problem this way? >> Is there a more straightforward approach? > I think it's easier to use the fact that every C^1 function > is BV on bounded intervals; the proof is immediatefrom > the mean value theorem. Perhaps but i was thinking ahead and wondering what if i wish to prove it for [0,oo)? The bounded interval is then not an option but i can still piece it up. What i was aiming at was to question my choice of (+/-)-functions. Is there a more general way to pick them? Any comments at all? V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Provinf bounded variation >> I wish to prove that a function is of bounded variation. >> The function i'll regard is f(x) = cos(kx) and i have the >> following suggestion. >> f in BV([a,b]) if it can be written as a sum of two >> functions; one non-decreasing and one non-increasing. >> On the segments where cosine increases: >> +: cos(kx) -: 0 >> while where cosine decreases: >> +: -cos(kx) -: 2cos(kx) >> Is it right to regard the problem this way? >> Is there a more straightforward approach? > I think it's easier to use the fact that every C^1 function > is BV on bounded intervals; the proof is immediatefrom > the mean value theorem. > Perhaps but i was thinking ahead and wondering what if i > wish to prove it for [0,oo)? The bounded interval is then > not an option but i can still piece it up. What i was aiming > at was to question my choice of (+/-)-functions. But then the result is false, ie, piecing it up won't give BV on an unbounded interval. === Subject: Re: Provinf bounded variation > I wish to prove that a function is of bounded variation. > The function i'll regard is f(x) = cos(kx) and i have the > following suggestion. > f in BV([a,b]) if it can be written as a sum of two > functions; one non-decreasing and one non-increasing. >[...] > Is it right to regard the problem this way? > Is there a more straightforward approach? yes > (Only yes or no, please. It's sort of a homework.) === Subject: Re: Provinf bounded variation >> Is it right to regard the problem this way? >> Is there a more straightforward approach? > yes Was it ment to be a yes as in [yes, yes] or is it [yes, _]? I actually asked two questions. Also, given that the approach applied by me is correct, how could it be simplified? (That's not part of the assigment so it can be discussed freely - provided of course there is anything to say, hehe). -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Re: Provinf bounded variation I mean: >> Is there a more straightforward approach? > yes === Subject: Re: Provinf bounded variation > I mean: >> Is there a more straightforward approach? > yes you care to elaborate? Am i using a non-so-efficient theorem? Or can other functions be used? -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sence to be lazy --------------------------------------------------- === Subject: Topological Groups Somehow the two problems below seem related in that they ask for a continous real function rooted about 0. Let G be a regular topological group. But wait, I've repeated myself as all topological groups are regular. Anyway, G may be T1 if wished. For all closed K with 0 not in K, find a continuous function f:G -> R with f(0) = 0, f(K) = {1}. Yes, topological groups are uniform spaces and uniform spaces are Tychonov, hence f is apparent. Directly using the group structure, what suggestions have you for constructing such a function? Let G be 1st countable topological group. Find a continuous function n:G -> R with n(x) = 0 iff x = 0 n(-x) = n(x) n(x + y) <= n(x) + n(y) Directly, for all x, 0 <= n(x). By continuity, it would follow for all r > 0, { x : n(x) < r } is open. ---- === Subject: a finite group and embedding in GL_n(Z) I have probably a very easy question and i have proof which i don't too sure to be ok. Here is it: Let G be a finite group of order n. Is it true that G can be embedded in a group GL_n(Z). ? My idea: Yes, it is true. By Cayley's theorem G can be embedded in S_n - the permutation group of order n!. Now we can associate with each element sigma of this isomorphic with G subgroup G' of S_n a permutation matrix P such that P_{ij} = 1 if sigma(i) = j. So as G' form a subgroup these obtained from elents from G' permutation matrices will also form a subgroup in P_n and as P_n is a subgroup of GL_n(Z) we will have this image of G as a sungroup of GL_n(Z). Is it ok? Sorry if i explained it incorrectly. === Subject: Re: a finite group and embedding in GL_n(Z) Your proof is correct. Philippe === Subject: Some examples in elementary ergodic theory. Hi everydody, I have two questions, which are related to Kolmogorov (or measure-theoretic, or metric) entropy in ergodic theory. 1Á) Someone knows an example of a measurable dynamical system, whose measure-theoretic entropy is infinite ? 2Á) If we consider the maps f : x -> mu * x * (1 - x), with mu such that f([0;1]) subset [0;1]. What is an invariant measure for this map, and can you compute its metric entropy ? (Furthermore, what is a measure such that the measure-theoretic entropic h(f, mu) is maximal ?) Philippe. === Subject: Re: Another Reason Why Feinstein's Proof is Incorrect # On 1 Jun 2006 09:18:54 -0700, Craig Feinstein # #>But I might as well respond now as it appears that #>more than one person believes it and I may be able to help some people #>improve their critical thinking skills by changing their minds: #>My proof never claims that you have to treat every integer #>individually. My proof says: let's pretend that we have a proof of #>Collatz with L bits. It then shows that there is a specific n for which #>any proof that Collatz halts at one with input n requires at least L+1 #>bits. # # As I've pointed out several times, this is exactly where the # error is. Or rather where the essential error is - the other # problems about how you haven't specified this or defined that # can't be ignored but they can be fixed. # # What follows is not that the proof the Collatz halts with # input n takes at least L+1 bits. What follows is that that # proof, _including_ a specification of what n you're talking # about, takes at least L+1 bits. # # And now (surprise) the contradiction goes away. I think that's the most beautiful mistake in Feinstein's paper. --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: Re: Another Reason Why Feinstein's Proof is Incorrect <2s5082tmjrsci3s7tnpcfs5270b172qd65@4ax.com> <448c08d7$0$8024$3b214f66@aconews.univie.ac.at # On 1 Jun 2006 09:18:54 -0700, Craig Feinstein But I might as well respond now as it appears that > #>more than one person believes it and I may be able to help some people > #>improve their critical thinking skills by changing their minds: > # #>My proof never claims that you have to treat every integer > #>individually. My proof says: let's pretend that we have a proof of > #>Collatz with L bits. It then shows that there is a specific n for which > #>any proof that Collatz halts at one with input n requires at least L+1 > #>bits. > # As I've pointed out several times, this is exactly where the > # error is. Or rather where the essential error is - the other > # problems about how you haven't specified this or defined that > # can't be ignored but they can be fixed. > # What follows is not that the proof the Collatz halts with > # input n takes at least L+1 bits. What follows is that that > # proof, _including_ a specification of what n you're talking > # about, takes at least L+1 bits. > # And now (surprise) the contradiction goes away. > I think that's the most beautiful mistake in Feinstein's paper. > --Gerhard I don't get this. In my experience David Ullrich is usually right, so presumably I am missing something. I don't see where in Feinstein's paper these L+1 bits have to include a specification of what n you're talking about. As far as I can see, these L+1 bits are n mod 2, T(n) mod 2, T^2(n) mod 2 etc. These bits may indirectly *imply* the value of n, but I don't see that the value of n is literally specified. Can anyone explain? (BTW, I am not defending the paper. I think it is either completely wrong or makes a trivial observation in a very complicated way, depending on what types of proof are allowed. I just don't understand this particular comment.) === Subject: Example+Fourier transform Could you give examples of f in L^2(R) - L^1(R) such that the fourier transform of f is in L^1(R). Under what circumstances can this happen ? === Subject: Re: Example+Fourier transform >Could you give examples of f in L^2(R) - L^1(R) such that the fourier >transform of f is in L^1(R). Under what circumstances can this happen ? Try sin(x)/x. This is always the case when f is the Fourier transform of a discontinuous L^1 function, as the Fourier transform of an L^1 function is continuous. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Example+Fourier transform > Could you give examples of f in L^2(R) - L^1(R) such that the fourier > transform of f is in L^1(R). sin(x)/x is a standard example; it's FT is a constant times the characteristic function of an interval, hence is in L^everything. > Under what circumstances can this happen ? Dunno. === Subject: Homomorphism I've been trying to find a good definition of homomorphism which is both general and precise. The definition is stated in many different ways by different authors. I believe this is very close to the essential definition. A mapping between two sets, which preserves the algebraic structure of the domain set. i.e. if a particular algebraic relation is true in the domain, it will be true after the mapping has been applied, in the range. A homomorphism between groups G and H is a map f that preserves the group operation, i.e. if p, q are in G then f(pq)=f(p)f(q). I want to say something like given a ???? (S,*) and a ???? (T,@), a mapping f:S->T such that for all s1,s2 in S, f(s1)=t1, f(s2)=t2, and f(s1*s2)=t1@t2. Is there a formal and general term for a set upon which a binary operation is defined? IOW, is there term I can use to replace the '????'? S and T are understood to be sets. Likewise * and @ are understood to be binary connectives. -- Nil conscire sibi === Subject: Re: Homomorphism days. My association with the Department is that of an alumnus. >I've been trying to find a good definition of homomorphism which is both >general and precise. The definition is stated in many different ways by >different authors. If you want to stick to ->algebraic<- homomorphisms, you are on the right track. What you are grasping for is the concept of universal algebra. However, homomorphism is not restricted to algebras. In particular, not restricted to sets equipped with a structure. That point of view, the element-free view, lies in Category Theory, and the work of Lawvere and others is relevant there. >I want to say something like given a ???? (S,*) and a ???? (T,@), a mapping >f:S->T such that for all s1,s2 in S, f(s1)=t1, f(s2)=t2, and >f(s1*s2)=t1@t2. Is there a formal and general term for a set upon which a >binary operation is defined? IOW, is there term I can use to replace the >'????'? S and T are understood to be sets. Likewise * and @ are understood >to be binary connectives. And why just binary? (-; Okay. Let's stick to algebras, i.e., a homomorphism will be a map of underlying sets. Given a set S, and an ordinal a, an a-ary operation on S is a map S^a->S. Thus, a binary operation is a map S^2->S; a ternary operation is a map S^3->S; an omega-ary operation is a map from S^{omega}->S; a unary operation is a map S->S. Another special kind is the zero-ary operations or nullary operations. These are maps from S^0 -> S. Since S^0 is the set of all maps from the empty set to S, the only element of S^0 is the empty set. That is, S^0 = {emptyset}. Thus, a zero-ary operation is a map from a singleton to S. This is equivalent to specifying an element of S. Thus, a nullary operation is equivalent to specifying an element. A type Omega will be an ordered pair, Omega=(W,ari_W), where W is a set, and ari_W is a map from W to the ordinals. An element s of W is an operation symbol, and for each such s, ari_W(s) is the arity of the operation symbol s. Thus, a type Omega represents a collection of operations, together with their arity. If Omega=(W,ari_W) is a type, then an algebra of type Omega is a pair, (A,{s_A}_{s in W}), where A is a set, and for each s in W, s_A is an ari_W(s)-ary operation on A. For example: take Omega to consist of three operations, e, *, and -, with corresponding arities 0, 2, and 1. That is, e is a nullary operation; * is a binary operation, and - is a unary operation. Call them, in a complete abusive use of notation, neutral, multiplication, and inversion. An algebra of type Omega will be a set A, together with three operations: a nullary operation e_A (the neutral of A), *_A (multiplication of A), and -_A (inversion of A). We also say A is an Omega-algebra. For example, a group G will be an algebra of type Omega, with e_G being the function that maps {emptyset} to the identity element of G; *_G being the group operation of G (which is a map from G x G to G); and -_G being the operation of taking inverses in G (the map from G to G that sends g to g^{-1}). Every group is an algebra of this type, but not every algebra of this type is a group (because a group has to satisfy a bunch of ancillary conditions). Another examples: if R is a ring, then a left R-module can be described as a certain type of Omega-algebra, where in this case Omega consists of a binary operation +, a unary operation -, a nullary operation called 0, and one operation symbol r for each element r of R, corresponding to a unary operation. + corresponds to the module addition, - to taking the additive inverse, 0 to the neutral element, and for each r in R, the unary operation that sends m to rm. Once you have the notion of types, it is easy to describe just what a homomorphism should be: DEF. Given a type Omega=(W,ari), and two Omega-algebras A and B, a HOMOMORPHISM f:A->B will be a set map from the underlying set of A to the underlying set of B, such that for each s in W, and each ari(s)-tuple (a_i) of elements of A, f( s_A(a_i)) = s_B(f(a_i)). Note that by definition s_A and s_B are both ari(s)-ary operations, so if (a_i) is an ari(s)-tuple of elements of A, then (f(a_i)) is an ari(s)-tuple of elements of B, and it makes sense to compue s_B(f(a_i)). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Homomorphism >I've been trying to find a good definition of homomorphism which is both >general and precise. The definition is stated in many different ways by >different authors. I believe this is very close to the essential >definition. >url='http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=2487' A mapping between two sets, which preserves the algebraic structure of the >domain set. i.e. if a particular algebraic relation is true in the domain, >it will be true after the mapping has been applied, in the range. > A homomorphism between groups G and H is a map f that preserves the group >operation, i.e. if p, q are in G then f(pq)=f(p)f(q). > That's not nearly as well put as it ought to be, given that Cambridge University Press apparently sponsors that site. Even accepting the looseness of A mapping between two *sets* (my emphasis), and the implicit assumption that algebraic structure is the only kind of structure about which one ought to speak of structure-preserving maps as homomorphisms, I still think it's very bad not to emphasize from the start that *both* the domain set and the range set have to be equipped with the *same kind* of algebraic structure. Further, the example given is (thank goodness) correct, but deceptively special, since in fact the algebraic structure of a group includes not just the binary group operation of multiplication but also a nullary group operation (the identity element) and a unary group operation (inversion); it happens to be true *for groups*, because of the nature of the group axioms, that if a map between the underlying sets G and H of two groups preserves the binary operation then it also preserves the nullary and unary operations, but that is not true for all types of algebraic structures, including many of genuine interest (i.e., not merely concocted to keep people honest). However, let us put aside these massive quibbles, and proceed. >I want to say something like given a ???? (S,*) and a ???? (T,@), (I am glad to see that, uncorrupted by thesaurus.maths.org, you have explicitly required both domain and range to be ????s, and explicitly noted that they are *structured* sets) >a mapping >f:S->T such that for all s1,s2 in S, f(s1)=t1, f(s2)=t2, and >f(s1*s2)=t1@t2. In my opinion, it's bad style to introduce t1 and t2 by name when you could simply write such that if s1 and s2 are in S then f(s1*s2) = f(s1)@f(s2) (or the like). However, that is a nearly massless quibble, so forget it by all means. >Is there a formal and general term for a set upon which a >binary operation is defined? IOW, is there term I can use to replace the >'????'? S and T are understood to be sets. Likewise * and @ are understood >to be binary connectives. You are (re)developing universal algebra, and if you are both good and lucky, Alberto Magidin may explain it to you. Alternatively, james dolan may explain why you should be (re)developing category theory instead (and perhaps *not* burdening understanding--your or others'--with the requirement that S and T are sets). Lee Rudolph === Subject: Re: Homomorphism days. My association with the Department is that of an alumnus. [.snip.] >You are (re)developing universal algebra, and if you are both good >and lucky, Alberto Magidin may explain it to you. Actually, my cousin Beto is a movie director so I doubt he will try. And Magidin is his second last name (his mother's maiden name). (-; -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Membership in a polynomial ring z^2/w is or is not in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z). I've tried to algebraically transform z^2/w with the relations xw=yz, z^3=yw^2 and y^3=x^2z, but I haven't been able to find a polynomial f so that f=z^2/w (mod xw=yz, z^3=yw^2, y^3=x^2z) yet. So, I suspect that z^2/w is not in that ring. However, I don't know how to prove that yet. Please help. -kira === Subject: Re: Membership in a polynomial ring Oh, I forgot: the field k is assumed algebraically closed. > z^2/w > is or is not in the ring > k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z). > I've tried to algebraically transform z^2/w with the relations xw=yz, > z^3=yw^2 and y^3=x^2z, but I haven't been able to find a polynomial f so > that > f=z^2/w (mod xw=yz, z^3=yw^2, y^3=x^2z) > yet. So, I suspect that z^2/w is not in that ring. > However, I don't know how to prove that yet. > Please help. > -kira === Subject: Re: Membership in a polynomial ring z^2/w > is or is not in the ring > k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z). > I've tried to algebraically transform z^2/w with the relations xw=yz, > z^3=yw^2 and y^3=x^2z, but I haven't been able to find a polynomial f so > that > f=z^2/w (mod xw=yz, z^3=yw^2, y^3=x^2z) > yet. So, I suspect that z^2/w is not in that ring. > However, I don't know how to prove that yet. > Please help. > -kira z^2/w is not in the ring. Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) and f(x,y,z,w) * w - z^2 is zero in this ring. So there are polynoms f1, f2, f3, f4: f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + f3(x,y,z,w)*(y^3 - x^2z) Let w = 0. -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - x^2z) Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all monoms in f1(x,y,z,0)*(-yz) are multiple of yz. Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all monoms in f1(x,y,z,0)*(-yz) are multiple of z^3. And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) because all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. So such f doesn't exist, and z^2/w is not in the ring. === Subject: Re: Membership in a polynomial ring Sorry, but I don't get something here in your solution: > z^2/w is not in the ring. > Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) > and f(x,y,z,w) * w - z^2 is zero in this ring. > So there are polynoms f1, f2, f3, f4: > f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + > f3(x,y,z,w)*(y^3 - x^2z) > Let w = 0. (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - x^2z) Why is it that z^2 must show up as one of the terms on the right side of (1)? By the relation z^3 = yw^2 , if w=0, then z=0 also. So, -z^2 = 0 and the left side of (1) is 0. Also, by relation y^3 =x^2 z and since z=0, we must have y=0. So the right side of (1) is also 0, and there is no contradiction. > Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all > monoms in > f1(x,y,z,0)*(-yz) are multiple of yz. > Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all > monoms in > f1(x,y,z,0)*(-yz) are multiple of z^3. > And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) > because > all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. > So such f doesn't exist, and z^2/w is not in the ring. === Subject: Re: Membership in a polynomial ring z^2/w is not in the ring. > Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) > and f(x,y,z,w) * w - z^2 is zero in this ring. > So there are polynoms f1, f2, f3, f4: > f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + > f3(x,y,z,w)*(y^3 - x^2z) > Let w = 0. > (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - x^2z) > Why is it that z^2 must show up as one of the terms on the right side of > (1)? > By the relation z^3 = yw^2 , if w=0, then z=0 also. So, -z^2 = 0 and > the left side of (1) is 0. Also, by relation y^3 =x^2 z and since z=0, > we must have y=0. So the right side of (1) is also 0, and there is no > contradiction. > Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all > monoms in > f1(x,y,z,0)*(-yz) are multiple of yz. > Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all > monoms in > f1(x,y,z,0)*(-yz) are multiple of z^3. > And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) > because > all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. > So such f doesn't exist, and z^2/w is not in the ring. One polynom is equal to another iff all monoms in them are respectively equal. Monom z^2 is in left part of equation (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - x^2z) so it must be in the right part. === Subject: Re: Membership in a polynomial ring >>Sorry, but I don't get something here in your solution: >z^2/w is not in the ring. >Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) >and f(x,y,z,w) * w - z^2 is zero in this ring. >So there are polynoms f1, f2, f3, f4: >f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + >f3(x,y,z,w)*(y^3 - x^2z) >Let w = 0. >>(1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - x^2z) >>Why is it that z^2 must show up as one of the terms on the right side of >>(1)? >>By the relation z^3 = yw^2 , if w=0, then z=0 also. So, -z^2 = 0 and >>the left side of (1) is 0. Also, by relation y^3 =x^2 z and since z=0, >>we must have y=0. So the right side of (1) is also 0, and there is no >>contradiction. >Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all >monoms in >f1(x,y,z,0)*(-yz) are multiple of yz. >Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all >monoms in >f1(x,y,z,0)*(-yz) are multiple of z^3. >And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) >because >all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. >So such f doesn't exist, and z^2/w is not in the ring. > One polynom is equal to another iff all monoms in them are respectively > equal. > Monom z^2 is in left part of equation > (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - > x^2z) > so it must be in the right part. However, if you look at the relation xw = yz, then the polynomial xw is equal to yz even though they are not the same syntactically. This is obviously due to the nonunqiue factorization of this ring. So, how do you know that z^2 does not have another factorization that is one of the monomials on the other side of (1)? -kira === Subject: Re: Membership in a polynomial ring > Sorry, but I don't get something here in your solution: >> z^2/w is not in the ring. >> Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) >> and f(x,y,z,w) * w - z^2 is zero in this ring. >> So there are polynoms f1, f2, f3, f4: >> f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + >> f3(x,y,z,w)*(y^3 - x^2z) >> Let w = 0. > (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - > x^2z) > Why is it that z^2 must show up as one of the terms on the right side of > (1)? > By the relation z^3 = yw^2 , if w=0, then z=0 also. So, -z^2 = 0 and > the left side of (1) is 0. Also, by relation y^3 =x^2 z and since z=0, > we must have y=0. So the right side of (1) is also 0, and there is no > contradiction. >> Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all >> monoms in >> f1(x,y,z,0)*(-yz) are multiple of yz. >> Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all >> monoms in >> f1(x,y,z,0)*(-yz) are multiple of z^3. >> And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) >> because >> all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. >> So such f doesn't exist, and z^2/w is not in the ring. >> One polynom is equal to another iff all monoms in them are respectively >> equal. >> Monom z^2 is in left part of equation >> (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - >> x^2z) >> so it must be in the right part. > However, if you look at the relation > xw = yz, > then the polynomial xw is equal to yz even though they are not the same > syntactically. This is obviously due to the nonunqiue factorization of > this ring. > So, how do you know that z^2 does not have another factorization that is > one of the monomials on the other side of (1)? > -kira Ahh... I see. You already included all the zero elements in (1). So, to answer my own question, the equation xw = yz should really be written as (2) xw - yz = f(xw-yz) + g(z^3-yw^2) + h(y^3-x^2z). And so, xw and yz are indeed one of the monomials on the right side of (2). Sorry for being slow... -kira === Subject: Re: Membership in a polynomial ring > Sorry, but I don't get something here in your solution: >> z^2/w is not in the ring. >> Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) >> and f(x,y,z,w) * w - z^2 is zero in this ring. >> So there are polynoms f1, f2, f3, f4: >> f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + >> f3(x,y,z,w)*(y^3 - x^2z) >> Let w = 0. > (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - > x^2z) >> Why is it that z^2 must show up as one of the terms on the right side of > (1)? >> By the relation z^3 = yw^2 , if w=0, then z=0 also. So, -z^2 = 0 and > the left side of (1) is 0. Also, by relation y^3 =x^2 z and since z=0, > we must have y=0. So the right side of (1) is also 0, and there is no > contradiction. >> Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all >> monoms in >> f1(x,y,z,0)*(-yz) are multiple of yz. >> Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all >> monoms in >> f1(x,y,z,0)*(-yz) are multiple of z^3. >> And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) >> because >> all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. >> So such f doesn't exist, and z^2/w is not in the ring. >> One polynom is equal to another iff all monoms in them are respectively >> equal. >> Monom z^2 is in left part of equation >> (1) -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - >> x^2z) >> so it must be in the right part. > However, if you look at the relation > xw = yz, > then the polynomial xw is equal to yz even though they are not the same > syntactically. This is obviously due to the nonunqiue factorization of > this ring. > So, how do you know that z^2 does not have another factorization that is > one of the monomials on the other side of (1)? > -kira > Ahh... I see. You already included all the zero elements in (1). > So, to answer my own question, the equation > xw = yz > should really be written as > (2) xw - yz = f(xw-yz) + g(z^3-yw^2) + h(y^3-x^2z). > And so, xw and yz are indeed one of the monomials on the right side of (2). > Sorry for being slow... > -kira Yes, you are right. Equation (1) is identity in the ring K[x,y,z], but not in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z). === Subject: Re: Membership in a polynomial ring Ah... I see. Initially, I got stuck at f*w-z^2 = 0. You provided the step of taking w=0. -kira >>Oh, I forgot: the field k is assumed algebraically closed. > z^2/w >is or is not in the ring > k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z). >I've tried to algebraically transform z^2/w with the relations xw=yz, >z^3=yw^2 and y^3=x^2z, but I haven't been able to find a polynomial f so >that > f=z^2/w (mod xw=yz, z^3=yw^2, y^3=x^2z) >yet. So, I suspect that z^2/w is not in that ring. >However, I don't know how to prove that yet. >Please help. >-kira > z^2/w is not in the ring. > Let f(x,y,z,w) is in the ring k[x,y,z,w]/(xw=yz, z^3=yw^2, y^3=x^2z) > and f(x,y,z,w) * w - z^2 is zero in this ring. > So there are polynoms f1, f2, f3, f4: > f*w - z^2 = f1(x,y,z,w)*(xw - yz) + f2(x,y,z,w)*(z^3 - yw^2) + > f3(x,y,z,w)*(y^3 - x^2z) > Let w = 0. > -z^2 = f1(x,y,z,0)*(-yz) + f2(x,y,z,0)*(z^3) + f3(x,y,z,0)*(y^3 - x^2z) > Monom z^2 can't be one of monoms of f1(x,y,z,0)*(-yz) because all > monoms in > f1(x,y,z,0)*(-yz) are multiple of yz. > Monom z^2 can't be one of monoms of f2(x,y,z,0)*(z^3) because all > monoms in > f1(x,y,z,0)*(-yz) are multiple of z^3. > And monom z^2 also can't be one of monoms of f3(x,y,z,0)*(y^3 - x^2z) > because > all monoms in f3(x,y,z,0)*(y^3 - x^2z) are multiple of y^3 or x^2z. > So such f doesn't exist, and z^2/w is not in the ring. === Subject: are operators on complete distributive lattices continuous? I am working on operators op: L -> L where L is an infinite complete lattice which is infinitly distributive. Now my question is, what do I have to require of the lattice, that all such operators are continuous (limit preserving), like it is known to be the case for finite lattices. Or isn't that possible at all? -- Tobias Matzner *** eMail uy9g@rz.uni-karlsruhe.de === Subject: Re: The list of all natural numbers don't exist > The list of all natural numbers don't exist > Assumption: The list of all natural numbers exists. > The following sketch shows the sequence of the natural numbers in a > mono-cipher representation: > XX > XXX > XXXX > Since any next line contains the next natural number starting from 1 > (X), every line under the first line contains a natural number. > Since there are infinite many natural numbers, there are infinite many > X in the first column. > The next sketch shows the sequence of natural numbers in which the > digits are numbered with their index numbers. The index numbers are in > mono-cipher representation too, written with the cipher 0 and in > vertical orientation. > 0000 . . . > X000 > XX00 > XXX0 > XXXX > The sequence of the vertical sets of 0 represents the sequence of > the natural numbers as the sequence of the horizontal sets of the > X represents the natural numbers. So, in both sequences hold the > same sentence: since no natural number contains infinite many digits no > horizontal set of X and no vertical set of 0 is infinite in > size. > In consequence, there must be more X in the first column as there > are 0 in any column and there must be more 0 in the first > line as there are X in any line. > Now we can conclude from the second sketch that there must be at least > two lines in sequence, which contains (at least one) X but no > 0 and there must be at least one column, which contains (at least > one) 0 but no X. > Else there are not infinitely many natural numbers or there are natural > numbers with infinitely many digits - witch is both impossible. > If there are two lines in sequence, let's call them o and p, which > contains X but no 0, there is a following line q containing > a natural number which do not have an index number for every digit. > This is impossible since every digit of every natural number must have > an index number. > If there is a column, which contains 0 but no X, the next > column contains an index number, which don't correspond to a digit of > a natural number. This is impossible since for every index number N > there exists at least one natural number with N digits. > In consequence, the list of all natural numbers can't exist. > The shown contradictions have a great impact on axiomatic set theory: > - The proof Cantor's about the existence of more real numbers than > natural numbers don't hold > - The existence of infinite sets is self contradicting > - ZFC is self contradicting > Albrecht Siegfried Storz > Mannheim, Germany > This is the translation of the first posting of the thread Die Liste > aller nat.9frlichen Zahlen existiert NICHT One day I said that. I gave an example which shows that there are infinitelly many different lists of natural numbers and thus there is no list of all natural numbers. of one number per second , for Aleph-0 number of seconds, then this would be different from the list of natural numbers starting from zero but writtin at a rate of two numbers per second for Aleph-0 number of seconds. However that was rejected by most people in that forum. Best, Zuhair === Subject: Re: The list of all natural numbers don't exist > One day I said that. I gave an example which shows > that there are > infinitelly many different lists of natural numbers > and thus there is > no list of all natural numbers. > from zero at a rate > of one number per second , for Aleph-0 number of > seconds, then this > would be different from the list of natural numbers > starting from zero > but writtin at a rate of two numbers per second for > Aleph-0 number of > seconds. > However that was rejected by most people in that > forum. > Best, > Zuhair In that sense ( natural numbers associated with differents states of time ) it can be proved that there exists at least a characterization of the fact of being a natural number prime or not which depends on the time. However I think that this has nothing to do about the existence or not of the mentioned list. Fernando. === Subject: Re: The list of all natural numbers don't exist > The list of all natural numbers don't exist > Assumption: The list of all natural numbers exists. > The following sketch shows the sequence of the natural numbers in a > mono-cipher representation: > X > XX > XXX > XXXX > . > . > . > Since any next line contains the next natural number starting from 1 > (X), every line under the first line contains a natural number. > Since there are infinite many natural numbers, there are infinite many > X in the first column. > The next sketch shows the sequence of natural numbers in which the > digits are numbered with their index numbers. The index numbers are in > mono-cipher representation too, written with the cipher 0 and in > vertical orientation. > 0000 . . . > X000 > XX00 > XXX0 > XXXX > . > . > . > The sequence of the vertical sets of 0 represents the sequence of > the natural numbers as the sequence of the horizontal sets of the > X represents the natural numbers. So, in both sequences hold the > same sentence: since no natural number contains infinite many digits no > horizontal set of X and no vertical set of 0 is infinite in > size. > In consequence, there must be more X in the first column as there > are 0 in any column and there must be more 0 in the first > line as there are X in any line. > Now we can conclude from the second sketch that there must be at least > two lines in sequence, which contains (at least one) X but no > 0 and there must be at least one column, which contains (at least > one) 0 but no X. > Else there are not infinitely many natural numbers or there are natural > numbers with infinitely many digits - witch is both impossible. > If there are two lines in sequence, let's call them o and p, which > contains X but no 0, there is a following line q containing > a natural number which do not have an index number for every digit. > This is impossible since every digit of every natural number must have > an index number. > If there is a column, which contains 0 but no X, the next > column contains an index number, which don't correspond to a digit of > a natural number. This is impossible since for every index number N > there exists at least one natural number with N digits. > In consequence, the list of all natural numbers can't exist. > The shown contradictions have a great impact on axiomatic set theory: > - The proof Cantor's about the existence of more real numbers than > natural numbers don't hold > - The existence of infinite sets is self contradicting > - ZFC is self contradicting > Albrecht Siegfried Storz > Mannheim, Germany > This is the translation of the first posting of the thread Die Liste > aller nat.9frlichen Zahlen existiert NICHT > One day I said that. I gave an example which shows that there are > infinitelly many different lists of natural numbers and thus there is > no list of all natural numbers. > of one number per second , for Aleph-0 number of seconds, then this > would be different from the list of natural numbers starting from zero > but writtin at a rate of two numbers per second for Aleph-0 number of > seconds. > However that was rejected by most people in that forum. Yes, just because both lists contain exactly the same elements in exaclty the same order most people in the forum came to the conclusion that they were the same. -William Hughes === Subject: Re: The list of all natural numbers don't exist > The list of all natural numbers don't exist > Assumption: The list of all natural numbers exists. > The following sketch shows the sequence of the natural numbers in a > mono-cipher representation: > X > XX > XXX > XXXX > . > . > . > Since any next line contains the next natural number starting from 1 > (X), every line under the first line contains a natural number. > Since there are infinite many natural numbers, there are infinite many > X in the first column. > The next sketch shows the sequence of natural numbers in which the > digits are numbered with their index numbers. The index numbers are in > mono-cipher representation too, written with the cipher 0 and in > vertical orientation. > 0000 . . . > X000 > XX00 > XXX0 > XXXX > . > . > . > The sequence of the vertical sets of 0 represents the sequence of > the natural numbers as the sequence of the horizontal sets of the > X represents the natural numbers. So, in both sequences hold the > same sentence: since no natural number contains infinite many digits no > horizontal set of X and no vertical set of 0 is infinite in > size. > In consequence, there must be more X in the first column as there > are 0 in any column and there must be more 0 in the first > line as there are X in any line. > Now we can conclude from the second sketch that there must be at least > two lines in sequence, which contains (at least one) X but no > 0 and there must be at least one column, which contains (at least > one) 0 but no X. > Else there are not infinitely many natural numbers or there are natural > numbers with infinitely many digits - witch is both impossible. > If there are two lines in sequence, let's call them o and p, which > contains X but no 0, there is a following line q containing > a natural number which do not have an index number for every digit. > This is impossible since every digit of every natural number must have > an index number. > If there is a column, which contains 0 but no X, the next > column contains an index number, which don't correspond to a digit of > a natural number. This is impossible since for every index number N > there exists at least one natural number with N digits. > In consequence, the list of all natural numbers can't exist. > The shown contradictions have a great impact on axiomatic set theory: > - The proof Cantor's about the existence of more real numbers than > natural numbers don't hold > - The existence of infinite sets is self contradicting > - ZFC is self contradicting > Albrecht Siegfried Storz > Mannheim, Germany > This is the translation of the first posting of the thread Die Liste > aller nat.9frlichen Zahlen existiert NICHT > One day I said that. I gave an example which shows that there are > infinitelly many different lists of natural numbers and thus there is > no list of all natural numbers. The fact that you claim there are infinitely many things which are not the set of all naturals does not in any way prohibit the existence of such a thing. > of one number per second , for Aleph-0 number of seconds, then this > would be different from the list of natural numbers starting from zero > but writtin at a rate of two numbers per second for Aleph-0 number of > seconds. Only in the fatigue of the writer. > However that was rejected by most people in that forum. Rightfully so! > Best, > Zuhair === Subject: Re: The list of all natural numbers don't exist > The list of all natural numbers don't exist > Assumption: The list of all natural numbers exists. > The following sketch shows the sequence of the natural numbers in a > mono-cipher representation: > X > XX > XXX > XXXX > . > . > . > Since any next line contains the next natural number starting from 1 > (X), every line under the first line contains a natural number. > Since there are infinite many natural numbers, there are infinite many > X in the first column. > The next sketch shows the sequence of natural numbers in which the > digits are numbered with their index numbers. The index numbers are in > mono-cipher representation too, written with the cipher 0 and in > vertical orientation. > 0000 . . . > X000 > XX00 > XXX0 > XXXX > . > . > . > The sequence of the vertical sets of 0 represents the sequence of > the natural numbers as the sequence of the horizontal sets of the > X represents the natural numbers. So, in both sequences hold the > same sentence: since no natural number contains infinite many digits no > horizontal set of X and no vertical set of 0 is infinite in > size. > In consequence, there must be more X in the first column as there > are 0 in any column and there must be more 0 in the first > line as there are X in any line. > Now we can conclude from the second sketch that there must be at least > two lines in sequence, which contains (at least one) X but no > 0 and there must be at least one column, which contains (at least > one) 0 but no X. > Else there are not infinitely many natural numbers or there are natural > numbers with infinitely many digits - witch is both impossible. > If there are two lines in sequence, let's call them o and p, which > contains X but no 0, there is a following line q containing > a natural number which do not have an index number for every digit. > This is impossible since every digit of every natural number must have > an index number. > If there is a column, which contains 0 but no X, the next > column contains an index number, which don't correspond to a digit of > a natural number. This is impossible since for every index number N > there exists at least one natural number with N digits. > In consequence, the list of all natural numbers can't exist. > The shown contradictions have a great impact on axiomatic set theory: > - The proof Cantor's about the existence of more real numbers than > natural numbers don't hold > - The existence of infinite sets is self contradicting > - ZFC is self contradicting > Albrecht Siegfried Storz > Mannheim, Germany > This is the translation of the first posting of the thread Die Liste > aller nat.9frlichen Zahlen existiert NICHT > One day I said that. I gave an example which shows that there are > infinitelly many different lists of natural numbers and thus there is > no list of all natural numbers. > of one number per second , for Aleph-0 number of seconds, then this > would be different from the list of natural numbers starting from zero > but writtin at a rate of two numbers per second for Aleph-0 number of > seconds. > However that was rejected by most people in that forum. > Best, > Zuhair What does writing have to do with this? If you want to talk about writing, address an adecuate group. Perhaps alt.writers.kooks would suit you. === Subject: Re: The list of all natural numbers don't exist > The list of all natural numbers don't exist. There are theories where the list of all natural numbers does not exist. Nevertheless the set of all natural does exist in these theories. Let N denote the set of natural numbers. And let N* denote the lists over N, that is all m-tuples with elements from N. Formally N* is defined as: N* = union_m N^m We can define a predicate contains: x in <=> exists i (xi=x) The following holds: ~exists L in N* forall n in N n in L The proof is easy: Assume to the contrary that forall n in N n in L0 and L0 in N*. Because L0 in N*, there must be an m with L0 in N^m. But then m+1 not in L0. A contradiction to forall n in N n in L0. Bye === Subject: Re: The list of all natural numbers don't exist > Let N denote the set of natural numbers. And let N* denote > the lists over N, that is all m-tuples with elements from > N. Formally N* is defined as: > N* = union_m N^m One can call the lists in N* also bounded lists. Because for all L in N* the max { x | x in L } does exist. It is also possible to define unbounded lists, or omega words, denoted by N^omega. For these it is possible that max { x | x in L } does not exist. For example if an unbounded list contains all natural numbers, then the max does not exist. Its really easy to construct an unbound list that contains all natural numbers. Namely set xi = i. Albrecht: In the same vaine that max does not exist, you cannot count the 0's and X's in your depiction of the unbounded list. But in a bounded list you can count the 0's and X's. But a bounded list cannot contain all the natural numbers. Tricky, isn't it? But why should ZFC & Co. collapse now? Bye === Subject: Re: The list of all natural numbers don't exist Hi > Because L0 in N*, there must be an m with > L0 in N^m. But then m+1 not in L0. A > contradiction to forall n in N n in L0. Oops, max { x | x in L0 } + 1 not in L0. Bye === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > The incredible triangel of set theorie with INF: > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > An initinite set has no endpoint. infinite? ok. > This triangle has two sides. No. I talk about a triangle with two infinite and one finite side. Do you know one? I don't. My triangle of natural numbers has three finite sides or ... what? > Go talk to Tony Orlow. He also likes to insist that an > initinite set must have an endpoint. You can > do him a big favour by translating unspecifiable into > german. Unbestimmbar? Another formulation of the antinomie of infinite sets. Assumption: Infinite sets exists. The set of the natural numbers is infinite. The set is well-orderable in it's natural order (1, 2, 3, 4, ...). The set is subscriptable in this well-ordered form. The set of the natural numbers has the cardinality aleph_0. Since the set contains aleph_0 (omega) elements there must be an element with the subscript aleph_0 (omega). In the well-ordered natural form (1, 2, 3, 4, ...), the element with the index aleph_0 (omega) is an infinite number ( and member of the set). There is no infinite natural number so we gain a contradiction. Consequence: The natural numbers can't be well-ordered in their natural order (bad!) or the set doesn't exist. I agree with the second consequence. Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. >>The >>proof shows that if f: X -> P(X) is any mapping, then there exists >>y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural > numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > The incredible triangel of set theorie with INF: > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > infinite? ok. > > This triangle has two sides. > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > My triangle of natural numbers has three finite sides or ... what? > > Go talk to Tony Orlow. He also likes to insist that an > initinite set must have an endpoint. You can > do him a big favour by translating unspecifiable into > german. > Unbestimmbar? > Another formulation of the antinomie of infinite sets. > Assumption: Infinite sets exists. > The set of the natural numbers is infinite. The set is well-orderable > in it's natural order (1, 2, 3, 4, ...). > For one thing, the set of naturals with its normal ordering is not > merely well-orderable, it is well-ordered, in the sense that every > non-empty subset has a smallest member. > The set is subscriptable in this well-ordered form. > I have no idea what subscriptible is supposed to mean. a_1, a_2, a_3, a_4, ... Every element has an index or subscribt. I'm not shure about the correct terminus technicus in english. > The set of the natural numbers has the cardinality aleph_0. > Since the set contains aleph_0 (omega) elements > That does not follow. To say that a set has a certain cardinality says > nothing about how many elements it has unless the set is known to be > finite and threrfore has the same nuber of elements as indicated by some > natural number. At most, having a cardinality only says the set has > the same number of elements as any other set with the same cardinality. Oh, the set of natural numbers doesn't have aleph_0 elements? > In the well-ordered natural > form (1, 2, 3, 4, ...), the element with the index aleph_0 (omega) is > an infinite number ( and member of the set). > Such a member of any set satisfying the Peano properties does not exist. Yes. That's why the infinite set of the natural numbers leads to a contradiction. Albrecht S. storz === Subject: Re: The list of all natural numbers don't exist > The set is subscriptable in this well-ordered form. > I have no idea what subscriptible is supposed to mean. > a_1, a_2, a_3, a_4, ... > Every element has an index or subscribt. I'm not shure about the > correct terminus technicus in english. If you mean that the natural number can be used as subscripts, that is the same as saying that one can construct functions having the set of naturals as domain. But that does not mean that such a domain is a ever a member of itself. > . > The set of the natural numbers has the cardinality aleph_0. > Since the set contains aleph_0 (omega) elements > That does not follow. To say that a set has a certain cardinality says > nothing about how many elements it has unless the set is known to be > finite and threrfore has the same nuber of elements as indicated by some > natural number. At most, having a cardinality only says the set has > the same number of elements as any other set with the same cardinality. > Oh, the set of natural numbers doesn't have aleph_0 elements? I have no idea where you get that peculiar misinterpretation of what I said. The cardinality of the set of naturals is not a member of the set of naturals > In the well-ordered natural > form (1, 2, 3, 4, ...), the element with the index aleph_0 (omega) is > an infinite number ( and member of the set). > Such a member of any set satisfying the Peano properties does not exist. > Yes. That's why the infinite set of the natural numbers leads to a > contradiction. What contradiction is that? I have seen none. I have seen a false claim that the set of naturals must contain its own cardinality, but no proof. === Subject: Re: The list of all natural numbers don't exist . > The set of the natural numbers has the cardinality aleph_0. > Since the set contains aleph_0 (omega) elements > > That does not follow. > Oh, the set of natural numbers doesn't have aleph_0 elements? > I have no idea where you get that peculiar misinterpretation of what I > said. ??? Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist I have seen a false claim that the set of naturals must contain its own > cardinality, but no proof. There is no universe in ZF. A universal set would contain itself. A variety of results about completeness of the naturals exhibit that the proof of some true statements about them require there being an infinite element or elements. The naturals are compact. Compactification with a one-point infinity or singularity is implicit. There are a variety of other reasonings, proofs, to that effect. Ross === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > The incredible triangel of set theorie with INF: > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > An initinite set has no endpoint. > infinite? ok. > This triangle has two sides. > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. You may talk about a triangle with two infinite and one finite side, but what you pictured above is a triangle with two sides. Since there is no third side there is no contradiction. > My triangle of natural numbers has three finite sides or ... what? > Go talk to Tony Orlow. He also likes to insist that an > initinite set must have an endpoint. You can > do him a big favour by translating unspecifiable into > german. > Unbestimmbar? X is unspecifiable means that you cannot use any property of X in a proof by contradiction that X does not exist. I don't think that unbestimmbar carries this conotation. > Another formulation of the antinomie of infinite sets. > Assumption: Infinite sets exists. > The set of the natural numbers is infinite. The set is well-orderable > in it's natural order (1, 2, 3, 4, ...). The set is subscriptable in > this well-ordered form. > The set of the natural numbers has the cardinality aleph_0. > Since the set contains aleph_0 (omega) elements No. The set of natural numbers has cardinality aleph_0. True The set of natural numbers has aleph_0 elements. False. Cardinality (you really should learn the definition) means number of elements only in the finite case. A set with cardinality aleph_0 need not contain an aleph_0 th element. -William Hughes === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > The incredible triangel of set theorie with INF: > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > infinite? ok. > > This triangle has two sides. > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. It has three sides like any good triangel. The third side you just can't see. > Since there is no third side there is no contradiction. > My triangle of natural numbers has three finite sides or ... what? > > Go talk to Tony Orlow. He also likes to insist that an > initinite set must have an endpoint. You can > do him a big favour by translating unspecifiable into > german. > Unbestimmbar? > X is unspecifiable means that you cannot use > any property of X in a proof by contradiction that X does not exist. > I don't think that unbestimmbar carries this conotation. Ok. I'm not familiar with the specific issue. I just had tried to translate a word. > Another formulation of the antinomie of infinite sets. > Assumption: Infinite sets exists. > The set of the natural numbers is infinite. The set is well-orderable > in it's natural order (1, 2, 3, 4, ...). The set is subscriptable in > this well-ordered form. > The set of the natural numbers has the cardinality aleph_0. > Since the set contains aleph_0 (omega) elements > No. > The set of natural numbers has cardinality aleph_0. True > The set of natural numbers has aleph_0 elements. False. Ok. Tell me how many elements it has. Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > > > The incredible triangel of set theorie with INF: > > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > > infinite? ok. > > This triangle has two sides. > > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > It has three sides like any good triangel. The third side you just > can't see. No. If there were a third side it would have to be the last line. Since there is no last line there is no third side. > Since there is no third side there is no contradiction. > My triangle of natural numbers has three finite sides or ... what? > > Go talk to Tony Orlow. He also likes to insist that an > initinite set must have an endpoint. You can > do him a big favour by translating unspecifiable into > german. > > Unbestimmbar? > X is unspecifiable means that you cannot use > any property of X in a proof by contradiction that X does not exist. > I don't think that unbestimmbar carries this conotation. > Ok. I'm not familiar with the specific issue. I just had tried to > translate a word. > > Another formulation of the antinomie of infinite sets. > > Assumption: Infinite sets exists. > > The set of the natural numbers is infinite. The set is well-orderable > in it's natural order (1, 2, 3, 4, ...). The set is subscriptable in > this well-ordered form. > The set of the natural numbers has the cardinality aleph_0. > Since the set contains aleph_0 (omega) elements > No. > The set of natural numbers has cardinality aleph_0. True > The set of natural numbers has aleph_0 elements. False. > Ok. Tell me how many elements it has. The concept number of elements does not apply to infinite sets. Please repeat 50 times a day cardinality does not mean number of elements. -William Hughes === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > > > The incredible triangel of set theorie with INF: > > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > > infinite? ok. > > > This triangle has two sides. > > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > It has three sides like any good triangel. The third side you just > can't see. > No. If there were a third side it would have to be the last line. > Since there is no last line there is no third side. Look at this: x x xx x xx xxx x xx x x xxxx x xx x x x x xxxxx x xx x x x x x x xxxxxx x xx x x x x x x x x xxxxxxx ... If you claim, there is an object with infinite many elements, in this case infinite many lines (actually, at once), how could the first column and the diagonal be infinite without infinite base-side? You claim, that there is an acute-angled angle with infinite sides (infinitely long) but no infinite distance between the sides - never and nowhere. On what depends it, if somewhat can be infinite or not. Is infinity only possible in north-south- but never in west-east- orientation? Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist > >>How so? Cantor's theorem merely says that |X| < |P(X)| for >>all X. The >>proof shows that if f: X -> P(X) is any mapping, then there >>exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly > exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all > natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations > of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than > unicorns. > > The assumption that there is a set of all natural numbers > does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite > sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like > to use a > sieve which should hold water and wondering about why it is > always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem > a new > assumption is necessary that there are sets of points which has > a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers > at once > (as a set) is like to say that there is a triangle with two > infinite > sides and a finite one. > > > > The incredible triangel of set theorie with INF: > > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > > infinite? ok. > > > This triangle has two sides. > > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > > It has three sides like any good triangel. The third side you just > can't see. > No. If there were a third side it would have to be the last line. > Since there is no last line there is no third side. > Look at this: > xx > xx > xxx > xx > x x > xxxx > xx > x x > x x > xxxxx > xx > x x > x x > x x > xxxxxx > xx > x x > x x > x x > x x > xxxxxxx > ... > If you claim, there is an object with infinite many elements, in this > case infinite many lines (actually, at once), how could the first > column and the diagonal be infinite without infinite base-side? Each of the triangles shown is finite, and no matter how far the series is continued each triangle will remain finite. If not, identify the point at which extending one more step jumps from a finite triangle to an infinite one. > You claim, that there is an acute-angled angle with infinite sides > (infinitely long) but no infinite distance between the sides - never > and nowhere. The distance between any point on one side and any point on the other is finite, however the set of all such distances has no finite upper bound any more than the lengths of the sides have a finite upper bound. There is a difference between a set containing objects with no finite upper bound on size and a set containing even one object of infinite size. === Subject: Re: The list of all natural numbers don't exist [SNIP] > There is a difference between a set containing objects with no finite > upper bound on size and a set containing even one object of infinite > size. This looks more like a case of set/element dyslexia: the belief that if EVERY ELEMENT of a set has some property then THE SET ITSELF must have that property. In this case, the property is is finite. That fallacy showed up clearly in the output of Herc a year or so ago. === Subject: Re: The list of all natural numbers don't exist > [SNIP] >> There is a difference between a set containing objects with no finite >> upper bound on size and a set containing even one object of infinite >> size. > This looks more like a case of set/element dyslexia: the belief that if > EVERY ELEMENT of a set has some property then THE SET ITSELF must have > that property. In this case, the property is is finite. Consider the following family of sets: N_1 = {0} N_2 = {0,1} N_3 = {0,1,2} ... N_i = {0,1,...,i-1} ... This family of sets has no upper bound. Let K_i = N_1 union N_2 union ... union N_i. It is not hard to prove that K_i = N_i. Note this is not a property of individual N_i, this is a property of any set that is the union of one or more N_i. Let K be the union of all N_i. I would argue this union is undefined because the family of sets, N_i, is unbounded. But, if K does exist, it must be equal to some N_i. This is property of K and not of any N_i. Russell - 2 many 2 count > That fallacy showed up clearly in the output of Herc a year or so ago. === Subject: Re: The list of all natural numbers don't exist > [SNIP] >> There is a difference between a set containing objects with no finite >> upper bound on size and a set containing even one object of infinite >> size. > This looks more like a case of set/element dyslexia: the belief that if > EVERY ELEMENT of a set has some property then THE SET ITSELF must have > that property. In this case, the property is is finite. > Consider the following family of sets: > N_1 = {0} > N_2 = {0,1} > N_3 = {0,1,2} > ... > N_i = {0,1,...,i-1} > ... > This family of sets has no upper bound. > Let K_i = N_1 union N_2 union ... union N_i. > It is not hard to prove that K_i = N_i. > Note this is not a property of individual N_i, > this is a property of any set that is the union of > one or more N_i. > Let K be the union of all N_i. > I would argue this union is undefined because > the family of sets, N_i, is unbounded. What has unboundedness to do with the existence of unions? And how does unboundedness disable the axiom that says for each set x there is a set y whose members are precisely the members of the members of x? > But, if K does exist, it must be equal to some N_i. Claimed but not proven. Only when Russell can prove his claims, will he have something, until then he has nothing. === Subject: Re: The list of all natural numbers don't exist > [SNIP] >> There is a difference between a set containing objects with no finite >> upper bound on size and a set containing even one object of infinite >> size. > This looks more like a case of set/element dyslexia: the belief that if > EVERY ELEMENT of a set has some property then THE SET ITSELF must have > that property. In this case, the property is is finite. > Consider the following family of sets: > N_1 = {0} > N_2 = {0,1} > N_3 = {0,1,2} > ... > N_i = {0,1,...,i-1} > ... > This family of sets has no upper bound. > Let K_i = N_1 union N_2 union ... union N_i. > It is not hard to prove that K_i = N_i. > Note this is not a property of individual N_i, > this is a property of any set that is the union of > one or more N_i. Actually, is a property of any set that is a union of N_1 through N_i for some i. > Let K be the union of all N_i. > I would argue this union is undefined because > the family of sets, N_i, is unbounded. > But, if K does exist, it must be equal to some N_i. No, because it is NOT a union of N_1 through N_i for some i. You're just begging the question here, where i stands in for the largest natural. Yes, if such an i existed then there would be a largest natural. Duh. > This is property of K and not of any N_i. > Russell > - 2 many 2 count > That fallacy showed up clearly in the output of Herc a year or so ago. === Subject: Re: The list of all natural numbers don't exist Hi > You claim, that there is an acute-angled angle with infinite sides > (infinitely long) but no infinite distance between the sides - never > and nowhere. On what depends it, if somewhat can be infinite or not. Is > infinity only possible in north-south- but never in west-east- > orientation? In projective geometry, you can map the plane onto a globe. Imagin the plane stretching from the north pole to all sides. On the globe your triangle will have finite length sides. Unfortunately this is not what one understands by the natural numbers. If we go along one side of this triangle, we will see that your X's are closer and closer. We can go omega steps, and we will never arrive the end of the triangle. Only if we take a step further, i.e. omega+1 steps, we will arrive at the end of the triangle. But we want the natural numbers to be omega and not omega+1. That is, we don't want to have an element c, such that: forall x(x=y) (2) Which contradicts the existence of c above. (Proof sketch: Take x=S(c), by (1) we have S(c)=c, a contradiction) Bye === Subject: Re: The list of all natural numbers don't exist Hi > Look at this: > xx > xx > xxx > xx > x x > xxxx > xx > x x > x x > xxxxx > xx > x x > x x > x x > xxxxxx > xx > x x > x x > x x > x x > xxxxxxx > ... Look instead at this: x x xx x xx x x xx x x x x xx x x x x x x xx x x x x x x x x xx x x x x x x x x x ... If you leave out the inside crosses, you should also do this for the last line in your initial segments. And then everything goes fine. So N becomes omega and not omega+1. Bye === Subject: Re: The list of all natural numbers don't exist <448D7420.5080600@fastmail.fm> <448D77AC.7050305@fastmail.fm Hi > Look at this: > x > x > xx > x > xx > xxx > x > xx > x x > xxxx > x > xx > x x > x x > xxxxx > x > xx > x x > x x > x x > xxxxxx > x > xx > x x > x x > x x > x x > xxxxxxx > ... > Look instead at this: > xx > xx > xx > x x > xx > x x > x x > xx > x x > x x > x x > xx > x x > x x > x x > x x > ... > If you leave out the inside crosses, > you should also do this for the last > line in your initial segments. You can't leave out everything. If ther are no crosses, there are space characters. > And > then everything goes fine. > So N becomes omega and not omega+1. If N is actual omega, there is an omegath element. Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist Am 12 Jun 2006 13:06:42 -0700 schrieb Albrecht: >> So N becomes omega and not omega+1. > If N is actual omega, there is an omegath element. Lieber Albrecht! Bist du so dumm das du nicht in ein paar simplen Formeln deinen Beweis hinschreiben kannst? Bilder sind da etwas l.8acherlich. Try to make a proof! A Picture is never a proof. -- Mit freundlichen Gr.9fssen Peter Nie¤en === Subject: Re: The list of all natural numbers don't exist Hi > If N is actual omega, there is an omegath element. Simply don't call your numbers natural numbers, call them natural numbers with a infinity point or what ever. Then define + and * please. And compare it with peano. Is it the same theory. Do the same consequences hold? Note: In peano one can simply define =< and > by: x= exists z(x+z=y) x>y :<=> ~x=y formally. But I assume a proof is possible. Bye === Subject: Re: The list of all natural numbers don't exist Hi > If N is actual omega, there is an omegath element. Yes outside N, but not inside N. I gave you the prove via Peano, that N~omega and not N~omega+1. Actually meghill states N~omega{0}. http://us.metamath.org/mpegif/df-ni.html And if you look there, its also possible to go another way, to derive Peano from ZFC. Indepently of that, either you have to accept Peano, and you arrive at N. Or you don't accept Peano and you arrive at something else. It seems that you don't accept Peano. So technically speaking you might not call your list a list of natural numbers, because not accepting Peano is not natural. So your list is a list of the members of omega+1. So what? You could also make a list of omega+omega. So when your triangles end, then new triangles appear: x xx x x ... x xx x x ... Or even depict omega^2. See for example: http://en.wikipedia.org/wiki/Image:Omega_squared.png But calling this natural is an abuse of the notion natural as it is used in connection with numbers in math. Although omega+omega, omega^2 etc.. are also natural, think for example of a mapping of omega+omega into the interval [0,2), derived from the mapping omega into the interval [0,1) by: n --> 1-1/(n+1). So its still naturally conceivable. But it is not what is meant by the term natural numbers. Bye === Subject: Re: The list of all natural numbers don't exist Hi > Hi > I gave you the prove via Peano, > that N~omega and not N~omega+1. > Actually meghill states N~omega{0}. > http://us.metamath.org/mpegif/df-ni.html Actually metamath is a little strange here. It defines N, as above, and NN as here: http://us.metamath.org/mpegif/df-n.html And they go all the way: Ord..>N ..> Z ..> Q ..> R ..> CC . v NN<.. ZZ<.. QQ<.. RR Probably to make the things comparable in the end. And to arrive at something like: http://us.metamath.org/mpegif/nthruc.html Hm. Scratching my head... bye === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > > > The incredible triangel of set theorie with INF: > > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > > infinite? ok. > > > This triangle has two sides. > > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > > It has three sides like any good triangel. The third side you just > can't see. > No. If there were a third side it would have to be the last line. > Since there is no last line there is no third side. > Look at this: > xx > xx > xxx > xx > x x > xxxx > xx > x x > x x > xxxxx > xx > x x > x x > x x > xxxxxx > xx > x x > x x > x x > x x > xxxxxxx > ... Yes in each of these *finite* diagrams there is a base side. This is simply another way of saying a finite set of natural numbers has a lagest element. The set of natural numbers does not have a largest element, so a diagram containing one line for each natural number does not have a base side. > If you claim, there is an object with infinite many elements, in this > case infinite many lines (actually, at once), how could the first > column and the diagonal be infinite without infinite base-side? Read carefully: THERE IS NO BASE SIDE > You claim, that there is an acute-angled angle with infinite sides > (infinitely long) but no infinite distance between the sides - never > and nowhere. Correct. This is just another way of saying that the set of natural numbers in infinite, but any given natural number is finite. True, the base side would have to be infinite, but this is not a problem because there is no base side. >On what depends it, if somewhat can be infinite or not. Is > infinity only possible in north-south- but never in west-east- > orientation? Correct. Any line that is not the base side will have finite extent. There is no base side. So every line has finite extent. -William Hughes === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > > > The incredible triangel of set theorie with INF: > > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > > infinite? ok. > > > This triangle has two sides. > > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > > It has three sides like any good triangel. The third side you just > can't see. > > No. If there were a third side it would have to be the last line. > Since there is no last line there is no third side. > Look at this: > x > x > xx > x > xx > xxx > x > xx > x x > xxxx > x > xx > x x > x x > xxxxx > x > xx > x x > x x > x x > xxxxxx > x > xx > x x > x x > x x > x x > xxxxxxx > ... > Yes in each of these *finite* diagrams there is a base side. > This is simply another way of saying a finite set of natural numbers > has a lagest element. The set of natural numbers does not have > a largest element, so a diagram containing one line for each > natural number does not have a base side. > If you claim, there is an object with infinite many elements, in this > case infinite many lines (actually, at once), how could the first > column and the diagonal be infinite without infinite base-side? > Read carefully: > THERE IS NO BASE SIDE Look: ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... x ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... xx x ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... x x xx x ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... x x x x x ... Now there is a (upper) base side which is infinite. Now, which of the finite sides can't be infinite. Two sides can be infinite as we have found before. The upper side is infinite. Will the left or the right side be finite in any case? > You claim, that there is an acute-angled angle with infinite sides > (infinitely long) but no infinite distance between the sides - never > and nowhere. > Correct. This is just another way of saying that the set of natural > numbers in infinite, but any given natural number is finite. > True, the base side would have to be infinite, but this is not a > problem > because there is no base side. See above. >On what depends it, if somewhat can be infinite or not. Is > infinity only possible in north-south- but never in west-east- > orientation? > Correct. Any line that is not the base side will have finite extent. > There is no base side. So every line has finite extent. You might see the connection to my first posting with infinite X and 0. Now you claim that there is no side infinite? So, there is no infinite set as I had shown. Albrecht S. Storz === Subject: Re: The list of all natural numbers don't exist >>How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >>proof shows that if f: X -> P(X) is any mapping, then there exists y in >>P(X) that is not in the range of f. >>The proof does not even mention lists. > > The cardinality of the power set of any set X strictly exceeds the > cardinality of X, said Tom listlessly. > > I have never questioned this part of Cantor's theorem. > It obviously applies to any finite set. > > I question whether there exists a set that contains all natural numbers. > The diagonal argument can be used to prove no set > contains every natural number. > (More precisely, no set that contains finite representations of > natural numbers can contain every finite representation of > a natural number.) > > If the set of all natural numbers does not exist, > then it is meaningless to say that the powerset of this set > is larger. > > That is like saying the powerset of unicorns is larger than unicorns. > > The assumption that there is a set of all natural numbers does > lead to contradictions. Unfortunately, many people convince > themselves these contradictions are properties of infinite sets. > > It is hard to convince someone a contradiction exists > if every time they see a contradiction they assume, > Oh, that is just another property of unicorns. > > And no, I am not impressed that the properties > of unicorns are consistent. > > > > Yes, you are right. To do Math with axiom of infinity is like to use a > sieve which should hold water and wondering about why it is always > dripping. > An example is the paradox of Banach-Tarski. To fix the problem a new > assumption is necessary that there are sets of points which has a > volume and there are sets of points without volume. > But no set of points has an extension (volume, length). > > The idea, that there are infinite many finite natural numbers at once > (as a set) is like to say that there is a triangle with two infinite > sides and a finite one. > > > > The incredible triangel of set theorie with INF: > > > x > xx > x x > x x > x x > x x > x x > x x > x x > x x > x x > x x > . . > . . > . . > > > > - infinite characters x in the first column > - infinite characters x on the diagonal > - finite space characters in any line > > An initinite set has no endpoint. > > infinite? ok. > > > This triangle has two sides. > > No. > I talk about a triangle with two infinite and one finite side. Do you > know one? I don't. > > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > > It has three sides like any good triangel. The third side you just > can't see. > > No. If there were a third side it would have to be the last line. > Since there is no last line there is no third side. > > Look at this: > > x > > x > xx > > x > xx > xxx > > x > xx > x x > xxxx > > x > xx > x x > x x > xxxxx > > x > xx > x x > x x > x x > xxxxxx > > x > xx > x x > x x > x x > x x > xxxxxxx > > ... > Yes in each of these *finite* diagrams there is a base side. > This is simply another way of saying a finite set of natural numbers > has a lagest element. The set of natural numbers does not have > a largest element, so a diagram containing one line for each > natural number does not have a base side. > > If you claim, there is an object with infinite many elements, in this > case infinite many lines (actually, at once), how could the first > column and the diagonal be infinite without infinite base-side? > Read carefully: > THERE IS NO BASE SIDE > Look: > ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... > x > ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... > xx > x > ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... > x x > xx > x > ... xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ... > x x > x x > x > ... > Now there is a (upper) base side which is infinite. And in the limit you get ...xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx... . . . . . . x x x x x Three infinite sides, no contradiction. But note that the first line does not correspond to any natural number. This is not the same as your original diagram where every line corresponded to a natural number. > Now, which of the finite sides can't be infinite. Two sides can be > infinite as we have found before. The upper side is infinite. Will the > left or the right side be finite in any case? > > You claim, that there is an acute-angled angle with infinite sides > (infinitely long) but no infinite distance between the sides - never > and nowhere. > Correct. This is just another way of saying that the set of natural > numbers in infinite, but any given natural number is finite. > True, the base side would have to be infinite, but this is not a > problem > because there is no base side. > See above. Above is something different. In you original diagram there is no base side. -William Hughes === Subject: Re: The list of all natural numbers don't exist sets. Please repeat 50 times a day cardinality does not mean > number of elements. Sez who? === Subject: Re: The list of all natural numbers don't exist sets. Please repeat 50 times a day cardinality does not mean > number of elements. > Sez who? Sez me. Cardinality is defined in terms of equivelence classes. For finite sets it turns out that cardinality and the naive concept of number of elements is the same. The naive concept of number of elements does not extend to infinite sets, cardinality does. So for infinite sets we have to use cardinality. If you want you can call the cardinality of an inifite set the number of elements in the set, as long as you realize that you are using a new concept of number of elments (e.g. it is now possible for a set to have the same number of elements as a proper subset), but applying the term number of elements to an infinite set frequently leads to confusion. -William Hughes === Subject: Re: The list of all natural numbers don't exist > You may talk about a triangle with two infinite and one finite side, > but what you pictured above is a triangle with two sides. > It has three sides like any good triangel. The third side you just > can't see. Not seeing what is not there to be seen is a good deal saner than imagining things that do not exist. === Subject: Re: The list of all natural numbers don't exist > ... > xxxx = four > > Are you claiming that four is not a natural number? > > No. I said that about the list of natural numbers, not about a finite > sublist > of the natural numbers. > The formula I described works on any list of natural numbers. > How does it work with an infinite list of natural numbers? There is no such thing. > More strict, it is easily proven that the diagonal number has > one more x than the last member of the list. This works well > with lists that have a last element, and the proof applies only to > lists that have a last member. The argument does not go through > with lists that do *not* have a last element. > It is more accurate to say the diagonal number has one more x > than the largest number in the list. > But a list of natural numbers has a maximal element when it is finite. And all lists of natural numbers are finite. > Of course, the argument above also shows that any list of natural > numbers > has a largest number. > Not at all. Where do you think the proof is flawed? The proof is almost identical to the normal diagonal argument. > I can provide other proofs in you want. > I am not interested in that. Pray show how your argument shows that any > list of natural numbers has a largest element. The only thing it shows > is that any finite list of natural numbers has a largest element. We did > know that already. Perhaps you should first provide a definition of these infinite sets you keep talking about. Every set of natural numbers has a largest member. Proof: Let: {} = zero {{}} = one {{},{{}}} = two ... zero union one = {} u {{}} = {{}} Given the members of a set, assume we can take the union of the members of that set. Given a set that contains only natural numbers, the union of the members of that set equals the largest natural number in the set. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist ... > The formula I described works on any list of natural numbers. > > How does it work with an infinite list of natural numbers? > There is no such thing. You state so. Can you give a proof? Do you know what the mathematical term list implies? > But a list of natural numbers has a maximal element when it is finite. > And all lists of natural numbers are finite. In that case there should be a largest list. What is that largest list? Consider the list of lists of natural numbers. By your reasoning that list must be finite. I am asking you to present the largest of all those lists. > Of course, the argument above also shows that any list of natural > numbers > has a largest number. > > Not at all. > Where do you think the proof is flawed? > The proof is almost identical to the normal diagonal argument. No. In the first place the normal diagonal argument does not assume that each and every list is finite. ... > Perhaps you should first provide a definition of these infinite sets > you keep talking about. > Every set of natural numbers has a largest member. What is the largest prime? The set of primes is a set of natural numbers. Or if it is not a set, what is the collection of all primes? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The list of all natural numbers don't exist > ... > The formula I described works on any list of natural numbers. > > How does it work with an infinite list of natural numbers? > There is no such thing. > You state so. Can you give a proof? Do you know what the mathematical > term list implies? > But a list of natural numbers has a maximal element when it is finite. > And all lists of natural numbers are finite. > In that case there should be a largest list. What is that largest list? > Consider the list of lists of natural numbers. By your reasoning that > list must be finite. I am asking you to present the largest of all those > lists. Just as there is no largest natural number, there is no largest list of natural numbers. If there was a largest list of natural numbers, it would be the list of all natural numbers. But, as this proof shows, there is no largest list. > Of course, the argument above also shows that any list of natural > numbers > has a largest number. > > Not at all. > Where do you think the proof is flawed? > The proof is almost identical to the normal diagonal argument. > No. In the first place the normal diagonal argument does not assume that > each and every list is finite. Neither do I. I make no assumptions about how many members are in the list. > ... > Perhaps you should first provide a definition of these infinite sets > you keep talking about. > Every set of natural numbers has a largest member. > What is the largest prime? The set of primes is a set of natural numbers. > Or if it is not a set, what is the collection of all primes? There is no set of all prime numbers. There might be a collection of all prime numbers. That would depend on what you mean by a collection. I have no problem with saying there are an infinite number of natural numbers or an infinite number of prime numbers. I object to the assumption that they can all be placed in a set. As you have pointed out, this means there is a largest set of natural numbers, and it is easy to prove that there can't be a largest set of natural numbers. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist > ... > The formula I described works on any list of natural numbers. > > How does it work with an infinite list of natural numbers? > > There is no such thing. > > You state so. Can you give a proof? Do you know what the mathematical > term list implies? ... > In that case there should be a largest list. What is that largest list? > Consider the list of lists of natural numbers. By your reasoning that > list must be finite. I am asking you to present the largest of all those > lists. > Just as there is no largest natural number, > there is no largest list of natural numbers. > If there was a largest list of natural numbers, > it would be the list of all natural numbers. > But, as this proof shows, there is no largest list. No, the proof does not show such a thing, because it starts with the assumption that a list is finite. > Of course, the argument above also shows that any list of natural > numbers > has a largest number. > > Not at all. > > Where do you think the proof is flawed? > The proof is almost identical to the normal diagonal argument. > > No. In the first place the normal diagonal argument does not assume that > each and every list is finite. > Neither do I. You do. What is the largerst element of an infinite list of natural numbers? And you reply there is no such list. Pray show where in the proof you have given that you show there is no infinite list of natural numbers. > I make no assumptions about how many members are in the list. You do. You assume the number of members is finite. > Perhaps you should first provide a definition of these infinite sets > you keep talking about. > Every set of natural numbers has a largest member. > > What is the largest prime? The set of primes is a set of natural numbers. > Or if it is not a set, what is the collection of all primes? > There is no set of all prime numbers. Again an assertion. Why is there no set of all prime numbers? > There might be a collection of all prime numbers. > That would depend on what you mean by a collection. Yes, and it also would depend on what you mean by set. How do you define a set? Apparently you do not use the mathematical definition. > I have no problem with saying there are an infinite number > of natural numbers or an infinite number of prime numbers. > I object to the assumption that they can all be placed in a set. Give your definition of set so that we know what we are talking about. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The list of all natural numbers don't exist What is the largest prime? The set of primes is a set of natural numbers. >> Or if it is not a set, what is the collection of all primes? > There is no set of all prime numbers. > There might be a collection of all prime numbers. > That would depend on what you mean by a collection. > I have no problem with saying there are an infinite number > of natural numbers or an infinite number of prime numbers. > I object to the assumption that they can all be placed in a set. > As you have pointed out, this means there is a largest set of > natural numbers, and it is easy to prove that there can't be > a largest set of natural numbers. Let: N_0 = {0} N_1 = {0, 1} N_2 = {0, 1, 2} ... N_i = {0, 1, 2, ..., i} ... Then N = N_1 U N_2 U N_3 U ... In other words, N is the union of all of the finite sets of naturals. Given this fairly common definition of N, using your set theory it should be easy to prove that it is not, in fact, the largest set of naturals. If your theory is correct, then there must exist at least one other set of naturals, call it L, that is larger than N. What is L? Which naturals are in L that are not in N? === Subject: Re: The list of all natural numbers don't exist > What is the largest prime? The set of primes is a set of natural > numbers. > Or if it is not a set, what is the collection of all primes? >> There is no set of all prime numbers. >> There might be a collection of all prime numbers. >> That would depend on what you mean by a collection. >> I have no problem with saying there are an infinite number >> of natural numbers or an infinite number of prime numbers. >> I object to the assumption that they can all be placed in a set. >> As you have pointed out, this means there is a largest set of >> natural numbers, and it is easy to prove that there can't be >> a largest set of natural numbers. > Let: > N_0 = {0} > N_1 = {0, 1} > N_2 = {0, 1, 2} > ... > N_i = {0, 1, 2, ..., i} > ... > Then > N = N_1 U N_2 U N_3 U ... > In other words, N is the union of all of the finite sets of naturals. > Given this fairly common definition of N, using your set theory it > should be easy to prove that it is not, in fact, the largest set of > naturals. N_1 U N_2 is finite. N_1 U N_2 U N_3 is finite. N_1 U N_2 U N_3 U N_4 is finite. ... For which i does this union stop being finite? > If your theory is correct, then there must exist at least one > other set of naturals, call it L, that is larger than N. What is L? > Which naturals are in L that are not in N? If N doesn't exist then we can say the set of all natural numbers is the empty set. Any set of naturals would contain naturals not in N. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist There is no set of all prime numbers. >> There might be a collection of all prime numbers. >> That would depend on what you mean by a collection. >> I have no problem with saying there are an infinite number >> of natural numbers or an infinite number of prime numbers. >> I object to the assumption that they can all be placed in a set. >> As you have pointed out, this means there is a largest set of >> natural numbers, and it is easy to prove that there can't be >> a largest set of natural numbers. >> Let: >> N_0 = {0} >> N_1 = {0, 1} >> N_2 = {0, 1, 2} >> ... >> N_i = {0, 1, 2, ..., i} >> ... >> Then >> N = N_1 U N_2 U N_3 U ... >> In other words, N is the union of all of the finite sets of naturals. >> Given this fairly common definition of N, using your set theory it >> should be easy to prove that it is not, in fact, the largest set of >> naturals. > N_1 U N_2 is finite. > N_1 U N_2 U N_3 is finite. > N_1 U N_2 U N_3 U N_4 is finite. > ... > For which i does this union stop being finite? Obviously, for any finite i, the union of sets _up to that point_ is finite. So, for no finite i is the union infinite, obviously. But I didn't say N was the union of finite sets up to a given finite i, did I? I said N was the union of _all_ of the finite subsets. Choose an i, and the union of all the finite subsets up to that point is N_i. Obviously N_i is finite. Now add the next subset for i+1, which is N_i+1. Now do this for all N_i. That's N. If you stop at any particular N_i, you obviously haven't got all the subsets (nor all the naturals), because you've omitted N_i+1. So there is no finite N_i where you can stop at, nor is there any finite value i where N_i stops being finite. >> If your theory is correct, then there must exist at least one >> other set of naturals, call it L, that is larger than N. What is L? >> Which naturals are in L that are not in N? > If N doesn't exist then we can say the set of all natural numbers is the > empty set. You might say that, but since the set of all naturals contains at least the natural 1, the rest of us would say it's just ludicrous to say it's an empty set. > Any set of naturals would contain naturals not in N. Any set of naturals would contain naturals that are not in the set of all naturals? How's that again? === Subject: Re: The list of all natural numbers don't exist Hi > I said N was the union of _all_ of the finite subsets. In set theory there are two possibilities to definie the union: a) x in union_i in I S_i :<=> exists i in I x in S_i b) x in union S :<=> exists s in S x in S In variant a) there is an index set I involved, so that we sum up the sets over this index set. The index set is somehow external. In variant b) there is not an explizit index set involved, instead the elements of a particular set are summed up. There is nothing external. In set theory definition b) is usually assumed. The if we have set theory with only finite sets, no infinite sets can emerge. The powerset of a finite set is infinite, and so is the union. But if you say you have N and finite sets, and you sum up over the index N these finite sets, then an infinite set may result. But the you are using definition a) and you make a mixin of N into your sets. Bye === Subject: Re: The list of all natural numbers don't exist Correction > The powerset of a finite set is infinite, and so is the union. The powerset of a finite set is FINITE, and so is the union. |P(S)| = 2^|S| === Subject: Re: The list of all natural numbers don't exist Hi > b) x in union S :<=> exists s in S x in S Correction: b) x in union S :<=> exists s in S x in s The last s schould be lower case. Bye === Subject: Re: The list of all natural numbers don't exist > What is the largest prime? The set of primes is a set of natural > numbers. > Or if it is not a set, what is the collection of all primes? >> There is no set of all prime numbers. >> There might be a collection of all prime numbers. >> That would depend on what you mean by a collection. >> I have no problem with saying there are an infinite number >> of natural numbers or an infinite number of prime numbers. >> I object to the assumption that they can all be placed in a set. >> As you have pointed out, this means there is a largest set of >> natural numbers, and it is easy to prove that there can't be >> a largest set of natural numbers. > Let: > N_0 = {0} > N_1 = {0, 1} > N_2 = {0, 1, 2} > ... > N_i = {0, 1, 2, ..., i} > ... > Then > N = N_1 U N_2 U N_3 U ... > In other words, N is the union of all of the finite sets of naturals. > Given this fairly common definition of N, using your set theory it > should be easy to prove that it is not, in fact, the largest set of > naturals. > N_1 U N_2 is finite. > N_1 U N_2 U N_3 is finite. > N_1 U N_2 U N_3 U N_4 is finite. > ... > For which i does this union stop being finite? Finite unions of finite sets are finite, but infinite unions of finite sets need not be. For every finite union, one can find a natural number representing the number of set being unioned. The union of all initial sets of naturals is not a finite union. Unless Russell can give the finite number of sets being unioned. Well, can you, punk? === Subject: Re: The list of all natural numbers don't exist >> What is the largest prime? The set of primes is a set of natural >> numbers. >> Or if it is not a set, what is the collection of all primes? >> There is no set of all prime numbers. > There might be a collection of all prime numbers. > That would depend on what you mean by a collection. >> I have no problem with saying there are an infinite number > of natural numbers or an infinite number of prime numbers. > I object to the assumption that they can all be placed in a set. > As you have pointed out, this means there is a largest set of > natural numbers, and it is easy to prove that there can't be > a largest set of natural numbers. >> Let: >> N_0 = {0} >> N_1 = {0, 1} >> N_2 = {0, 1, 2} >> ... >> N_i = {0, 1, 2, ..., i} >> ... >> Then >> N = N_1 U N_2 U N_3 U ... >> In other words, N is the union of all of the finite sets of naturals. >> Given this fairly common definition of N, using your set theory it >> should be easy to prove that it is not, in fact, the largest set of >> naturals. >> N_1 U N_2 is finite. >> N_1 U N_2 U N_3 is finite. >> N_1 U N_2 U N_3 U N_4 is finite. >> ... >> For which i does this union stop being finite? > Finite unions of finite sets are finite, but infinite unions of finite > sets need not be. This is a very interesting statement. In ZF without the Axiom of Infinity, the definition of finite set is set. All sets are finite without the Axiom of Infinity. This is precisely why we need the Axiom of Infinity. It is impossible to prove there exists a set that contains the empty set and is closed under the successor function without it. We have to assume such a set exists by adding the Axiom of Infinity. I think we all agree that my proof works fine in ZF without the Axiom of Infinity (AoI). Adding AoI seems to do a lot more than just guarantee a certain set exists. Now, it seems we have distinguish between finite unions and infinite unions. > For every finite union, one can find a natural number representing the > number of set being unioned. > The union of all initial sets of naturals is not a finite union. We can prove that the set guaranteed to exist by AoI can not be any of the (finite) sets that existed before we added AoI. To me this is already a contradiction. What justification do we have for changing the definition of union besides trying to avoid this obvious contradiction? Why isn't there an Axiom of Infinite Unions? Are we just assuming the rules for taking the union of sets has changed because we added AoI? Russell - Gay couples were responsible for 9/11 and the illegal immigrants have weapons of mass destruction. === Subject: Re: The list of all natural numbers don't exist > Finite unions of finite sets are finite, but infinite unions of finite > sets need not be. > This is a very interesting statement. > In ZF without the Axiom of Infinity, > the definition of finite set is set. > All sets are finite without the Axiom of Infinity. Not necessarily. It is only that without such an axiom we do not know and cannot prove that there are infinite sets while with it we do know and can prove that there are. > Adding AoI seems to do a lot more than just > guarantee a certain set exists. > Now, it seems we have distinguish between finite unions > and infinite unions. if we index the sets in a union, the a finite union has a finite index set and an infinite union has an infinite index set, so we are adding nothing beyond the possibility of having infinite sets. > For every finite union, one can find a natural number representing the > number of set being unioned. > The union of all initial sets of naturals is not a finite union. > We can prove that the set guaranteed to exist by AoI > can not be any of the (finite) sets that existed before we added AoI. We cannot prove that the set guaranteed by the AoI does not exist in a set theory without that axiom. It is only that we cannot prove it does exist without an AoI to guarantee that existence. A set theory without an AoI is ambiguous on the matter of whether infinite sets can exist unless it specifically forbids them. > To me this is already a contradiction. > What justification do we have for changing the definition of union > besides trying to avoid this obvious contradiction? No one has changed any definition of union. > Why isn't there an Axiom of Infinite Unions? None is needed. > Are we just assuming the rules for taking the union of sets > has changed because we added AoI? Wikipedia: In axiomatic set theory and the branches of logic, mathematics, and computer science that use it, the axiom of union is one of the axioms of Zermelo-Fraenkel set theory, stating that, for any set x there is a set y whose elements are precisely the elements of the elements of x. So for any infinite set whose members are themselves sets the union of those members is already defined without any changes in definition being necessary. === Subject: Re: The list of all natural numbers don't exist > N_1 U N_2 is finite. > N_1 U N_2 U N_3 is finite. > N_1 U N_2 U N_3 U N_4 is finite. > ... > For which i does this union stop being finite? >> Finite unions of finite sets are finite, but infinite unions of finite >> sets need not be. > This is a very interesting statement. > In ZF without the Axiom of Infinity, > the definition of finite set is set. No, that's not correct. In ZF without AI the existence of infinite sets is undecideable. We can't prove that they exist, and we can't prove that they don't exist. > All sets are finite without the Axiom of Infinity. What you are talking about is ZF - {AI} + {UF}, where UF is the axiom of universal finiteness: the assertion that for every x, there exists a natural number n such that |x| = n. > This is precisely why we need the Axiom of Infinity. > It is impossible to prove there exists a set > that contains the empty set and is closed under > the successor function without it. > We have to assume such a set exists by adding the Axiom of Infinity. > I think we all agree that my proof works fine > in ZF without the Axiom of Infinity (AoI). UF cannot be proved in ZF - {AI}. Try again. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: The list of all natural numbers don't exist >> N_1 U N_2 is finite. >> N_1 U N_2 U N_3 is finite. >> N_1 U N_2 U N_3 U N_4 is finite. >> ... >> For which i does this union stop being finite? > Finite unions of finite sets are finite, but infinite unions of finite > sets need not be. >> This is a very interesting statement. >> In ZF without the Axiom of Infinity, >> the definition of finite set is set. > No, that's not correct. In ZF without AI the existence of infinite sets > is undecideable. We can't prove that they exist, and we can't prove that > they don't exist. OK. See if I get this right. In ZF without AI, we can prove that some sets are finite. We do this by showing the set has a cardinality equal to some (finite?) ordinal. (Are all ordinals finite without AI?) But, there are sets where we can't prove the set has a finite cardinality. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist >> N_1 U N_2 is finite. >> N_1 U N_2 U N_3 is finite. >> N_1 U N_2 U N_3 U N_4 is finite. >> ... >> For which i does this union stop being finite? > Finite unions of finite sets are finite, but infinite unions of finite > sets need not be. >> This is a very interesting statement. >> In ZF without the Axiom of Infinity, >> the definition of finite set is set. > No, that's not correct. In ZF without AI the existence of infinite sets > is undecideable. We can't prove that they exist, and we can't prove that > they don't exist. > OK. > See if I get this right. > In ZF without AI, we can prove that some sets are finite. > We do this by showing the set has a cardinality equal > to some (finite?) ordinal. > (Are all ordinals finite without AI?) There is no way to tell. > But, there are sets where we can't prove the set > has a finite cardinality. In ZF sans AI, correct, at least without some additional axioms. > Russell > - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist We do this by showing the set has a cardinality equal > to some (finite?) ordinal. Correct. > (Are all ordinals finite without AI?) Without the axiom of infinity, neither the statement that all ordinals are finite nor the negation of the statement that all ordinals are finite are theorems. > But, there are sets where we can't prove the set > has a finite cardinality. There are sets where we can't prove... is too vague as it perhaps suggests pointing to at least one specific set and not being able to prove that it is not finite. A more exact way to put it is that we cannot prove that there do not exist sets that are not finite. Or, more simply, we cannot prove that there are no infinite sets. MoeBlee === Subject: Re: The list of all natural numbers don't exist We do this by showing the set has a cardinality equal > to some (finite?) ordinal. > Correct. > (Are all ordinals finite without AI?) > Without the axiom of infinity, neither the statement that all ordinals > are finite nor the negation of the statement that all ordinals are > finite are theorems. > But, there are sets where we can't prove the set > has a finite cardinality. > There are sets where we can't prove... is too vague as it perhaps > suggests pointing to at least one specific set and not being able to > prove that it is not finite. A more exact way to put it is that we > cannot prove that there do not exist sets that are not finite. Or, more > simply, we cannot prove that there are no infinite sets. > MoeBlee Hi Russell, MoeBlee, hey how's it going, Russell I think it's interesting to consider real numbers between zero and one defined in terms of some integer radices, the base of the numeric expansion, that are not generally considered: base one and a base infinity. In either case, in representation of the non-integer portion of a real number, by a power series on the right and a polynomial on the left in the radix as it is generally described, expansions, basically a list of real numbers between zero and one is the same as the list of natural integers. That's with some consideration of a smallest real number, and why such a thing exists, for example as is known in counterexamples to real analysis, and how in the description of such an item, for example the integral iota-multiples, there are contiguous real numbers on the continuum, basically with the indefiniteness property, defined in terms of some definite real number, where Dedekind/Cauchy or expansions are insufficient to describe all real numbers in their completeness, gaplessness, on the real number line. In base one, for example, unary, tally marks, a representation of a real number is offset some definite real number, simply zero, and represents at its conclusion, simply one. A similar representation happens in base infinity, where there is a distinct numeric symbol for each integer, as there may well be, then as there is some completed infinity, over all of them, for example differentials from the infinitesimal analysis, over all of them their sum is equal to one. MoeBlee, when you say that removal of the axiom of infinity from ZF would lead to undecideability of the finiteness of all sets of ZFC, I think you say that from the perception that the axioms of ZFC are independent and irredundant. There is no universe in ZFC, so there is no universal ordinal M to be a model of ZFC. If you think there is model theory, how do you confront the notion that the existence of a universal ordinal would be contradictory to ZFC? Are you thus working outside of ZFC automatically to work with ZFC? When you claim that infinity, the axiom, in ZFC is independent and not redundant, without it in the iterative universe, all the sets iterated from the empty set and successive powersets and unions and then combinatoric subsets and so forth are finite. If you claim to then quantify over all of the sets in the iterative universe, that leads to a contradiction, no? MfG, Ross === Subject: Re: The list of all natural numbers don't exist the definition of finite set is set. > All sets are finite without the Axiom of Infinity. No, without the axiom of infinity, you cannot prove that all sets are finite. Dropping the axiom of infinity only leaves open the question whether all sets are finite. To prove that all sets are finite, you must not just drop the axiom of infinity, but you must adopt an axiom that does imply that all sets are finite. > This is precisely why we need the Axiom of Infinity. > It is impossible to prove there exists a set > that contains the empty set and is closed under > the successor function without it. Correct. But, as I said, simply dropping the axiom of infinity does not provide that all sets are finite but rather just dropping the axiom of infinity leaves open the question whether all sets are finite. > We have to assume such a set exists by adding the Axiom of Infinity. > I think we all agree that my proof works fine > in ZF without the Axiom of Infinity (AoI). I didn't look over what you purport to be a proof, but if it is in ZF without the axiom of infinity, then you cannot assume that all sets are finite. If you want a theorem that all sets are finite, then you must not just drop the axiom of infinity but you must also add an axiom that implies that all sets are finite. > Adding AoI seems to do a lot more than just > guarantee a certain set exists. > Now, it seems we have distinguish between finite unions > and infinite unions. The axiom of infinity combines with the other axioms to produce theorems. > We can prove that the set guaranteed to exist by AoI > can not be any of the (finite) sets that existed before we added AoI. First, there is no single certain the set said to exist by the axiom of infinity. The axiom just says that there is at least one (without saying whether there is or is not more than one) successor-inductive set. However, combined with other axioms, we do have the result that the the axiom of infinity is equivalent with the statement that the set of finite ordinals (the set of natural numbers) exists. Second, of course we can prove that no successor-inductive set is finite. But it is not true that we can prove that the set of natural numbers does NOT exist per ZF without the axiom of infinity. > To me this is already a contradiction. A contradiction is a formula and its negation. You have shown no contradiction. > What justification do we have for changing the definition of union > besides trying to avoid this obvious contradiction? We never changed the definition. You have serious misunderstandings of the axiomatic method and methods of definition in axiomatic systems. The definition of 'union of x' is not changed by adopting the axiom of infinity. > Why isn't there an Axiom of Infinite Unions? Unless there is a need for one, it's not likely that one will be adopted. > Are we just assuming the rules for taking the union of sets > has changed because we added AoI? No. We don't change the definition just by adopting another axiom. Of course, adopting an additional axiom will provide that some unions have certain properties, such as being infinite, but that is not at all a change in the definition, since the definition does not mention any properties whatsover other than that of being a union. Would you please say what set theory and mathematical logic textbooks you reference for your understanding of set theory? Perhaps we can refer back to the sections you have misunderstood so that you can rid yourself of the fundamental misunderstanding you have. MoeBlee === Subject: Re: The list of all natural numbers don't exist P.S. When I say things such as all sets are finite, I'm referring simply to theorems of the the theory as we read off the formulas. I'm not saying anything about the properties of objects in models for the theory. So, when I say something like there exist infinite sets, I just mean that the formula that reads as there exists an x such that x is infinite is a member of the theory. === Subject: Re: The list of all natural numbers don't exist But a list of natural numbers has a maximal element when it is finite. >> And all lists of natural numbers are finite. >> In that case there should be a largest list. What is that largest list? >> Consider the list of lists of natural numbers. By your reasoning that >> list must be finite. I am asking you to present the largest of all those >> lists. > Just as there is no largest natural number, > there is no largest list of natural numbers. > If there was a largest list of natural numbers, > it would be the list of all natural numbers. > But, as this proof shows, there is no largest list. Then, as a consequence of your proof, there cannot be a list of all the natural numbers, because that list would be the largest list. Suppose I have a set K, where 1. 0 (a natural) is in K, and 2. if natural n is in K, then natural n+1 is also in K. Your set theory tells us that K is a finite set. Since all finite sets of naturals have a largest element (otherwise they would not be finite), what is K_max, the largest element in K? Since your set theory says that there does not exist a list (or set) of all the naturals, K cannot contain all the naturals. So which naturals are not in K? The answer to this is trivial if you can tell us what K_max is, since then K_max+1 could not be in K. === Subject: Re: The list of all natural numbers don't exist > But a list of natural numbers has a maximal element when it is finite. > And all lists of natural numbers are finite. > In that case there should be a largest list. What is that largest list? > Consider the list of lists of natural numbers. By your reasoning that > list must be finite. I am asking you to present the largest of all > those > lists. >> Just as there is no largest natural number, >> there is no largest list of natural numbers. >> If there was a largest list of natural numbers, >> it would be the list of all natural numbers. >> But, as this proof shows, there is no largest list. > Then, as a consequence of your proof, there cannot be a list of > all the natural numbers, because that list would be the largest > list. Yes. Since there can be no largest list of natural numbers, there can be no list of all natural numbers. > Suppose I have a set K, where > 1. 0 (a natural) is in K, > and > 2. if natural n is in K, then natural n+1 is also in K. > Your set theory tells us that K is a finite set. My proof shows no such set exists. I have no idea whether a set that doesn't exist is finite or not. > Since all > finite sets of naturals have a largest element (otherwise they > would not be finite), what is K_max, the largest element in K? You tell me. You are the one claiming to have the largest list of natural numbers. What is the largest member of K? > Since your set theory says that there does not exist a list > (or set) of all the naturals, K cannot contain all the naturals. > So which naturals are not in K? The answer to this is trivial > if you can tell us what K_max is, since then K_max+1 could not > be in K. I have provided two methods to find a number not in K. One can use the diagonal argument to find such a number, or one can take the union of all the natural numbers in K and then add 1. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist there can be no list of all natural numbers. >> Suppose I have a set K, where >> 1. 0 (a natural) is in K, >> and >> 2. if natural n is in K, then natural n+1 is also in K. >> Your set theory tells us that K is a finite set. > My proof shows no such set exists. > I have no idea whether a set that doesn't exist is finite or not. >> Since your set theory says that there does not exist a list >> (or set) of all the naturals, K cannot contain all the naturals. >> So which naturals are not in K? > I have provided two methods to find a number not in K. > One can use the diagonal argument to find such a number, > or one can take the union of all the natural numbers in K > and then add 1. Since K contains all the natural numbers, the union of all the natural numbers in K is simply K itself. By add 1 to this set I assume you add a natural number that is not already in the set. (If you mean add 1 to the largest natural in K, that's impossible, since there is no largest natural in K.) Since K contains all the natural numbers, which one do we add that got left out? Which natural number does your diagonal method produce that is not in the set of all natural numbers? === Subject: Re: The list of all natural numbers don't exist >> Since there can be no largest list of natural numbers, >> there can be no list of all natural numbers. > Suppose I have a set K, where > 1. 0 (a natural) is in K, > and > 2. if natural n is in K, then natural n+1 is also in K. > Your set theory tells us that K is a finite set. >> My proof shows no such set exists. >> I have no idea whether a set that doesn't exist is finite or not. > Since your set theory says that there does not exist a list > (or set) of all the naturals, K cannot contain all the naturals. > So which naturals are not in K? >> I have provided two methods to find a number not in K. >> One can use the diagonal argument to find such a number, >> or one can take the union of all the natural numbers in K >> and then add 1. > Since K contains all the natural numbers, the union of all the > natural numbers in K is simply K itself. I describe a method that creates a number not in the list and you reply that, since the list is assumed to contain every natural number, we can just add the number the diagonal method created to our list and still have the same list. I could use the same argument to disprove the diagonal proof of the uncountability of the reals. I have a countable list, R, that contains every real number. Even if the diagonal proof creates a number not in R, I can just add this number to R and R will still be countable. > By add 1 to this set > I assume you add a natural number that is not already in the set. > (If you mean add 1 to the largest natural in K, that's impossible, > since there is no largest natural in K.) If there is a largest set of natural numbers there must exist a largest natural number. Since there is no largest natural number, there can be no largest list of natural numbers, therefore there is no set of all natural numbers. > Since K contains all the natural numbers, which one do we add > that got left out? Which natural number does your diagonal method > produce that is not in the set of all natural numbers? Which real number does Cantor's proof create that is not in the list of all real numbers? The point of the proof is to show that there is at least one natural number missing in any list of naturals numbers. You can't just argue that your list contains all numbers and the number created by the diagonal argument is already in the list. The diagonal argument provably creates a number not in the original list. Changing the list is cheating. Consider the following family of sets: N_1 = {0} N_2 = {0,1} N_3 = {0,1,2} ... N_i = {0,1,..,i-1} Earlier, someone (not me) said that the union of this family of sets could be used as a definition of the set of all natural numbers. (If this was a valid definition of the set of all natural numbers, we would not need the Axiom of Infinity.) Let K_i be the union of: N_1 u N_2 u ... u N_i. Clearly, K_i is equal to N_i and this is true for all i. If the set of all natural numbers, N, equals the union of all N_i, then N must be equal to some N_i. We can modify Cantor's proof to get a similar result. For each N_i define a set M_i that contains every element of N_i and the next ordinal. N_1 = {0} M_1 = {0,1} N_2 = {0,1} M_2 = {0,1,2} ... For all i, the union of M_1 u M_2 u ... u M_i contains an element not in the union N_1 u N_2 u ... u N_i. The union of all M_i contains an element not in the union of all N_i. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist My proof shows no such set exists. >> I have no idea whether a set that doesn't exist is finite or not. >> Since your set theory says that there does not exist a list >> (or set) of all the naturals, K cannot contain all the naturals. >> So which naturals are not in K? >> I have provided two methods to find a number not in K. >> One can use the diagonal argument to find such a number, >> or one can take the union of all the natural numbers in K >> and then add 1. >> Since K contains all the natural numbers, the union of all the >> natural numbers in K is simply K itself. > I describe a method that creates a number not in the list > and you reply that, since the list is assumed to contain > every natural number, we can just add the number > the diagonal method created to our list and still have the same list. If the list contains all the naturals, how does your diagonal method produce a natural not in the list? > I could use the same argument to disprove the diagonal proof > of the uncountability of the reals. > I have a countable list, R, that contains every real number. > Even if the diagonal proof creates a number not in R, > I can just add this number to R and R will still be countable. Except that there can be no list of all reals. That's the difference. >> By add 1 to this set >> I assume you add a natural number that is not already in the set. >> (If you mean add 1 to the largest natural in K, that's impossible, >> since there is no largest natural in K.) > If there is a largest set of natural numbers there must exist > a largest natural number. Why? > Since there is no largest natural number, there can be no > largest list of natural numbers, therefore there is no set > of all natural numbers. So we can't fit all of the naturals into a single set, huh? What if we created a set of every natural except, say, 1? >> Since K contains all the natural numbers, which one do we add >> that got left out? Which natural number does your diagonal method >> produce that is not in the set of all natural numbers? > Which real number does Cantor's proof create that > is not in the list of all real numbers? Any list that is claimed to denumerate all of the reals is incomplete, because all the reals cannot be denumerated. So there are an infinite number of reals that are missing from any such list. That's why we say the list is uncountable. > The point of the proof is to show that there is at least > one natural number missing in any list of naturals numbers. > You can't just argue that your list contains all numbers > and the number created by the diagonal argument is > already in the list. If the list contains all the naturals, and your method produces a natural, then that natural is in the list. The naturals are countable, so we can say this. > The diagonal argument provably creates a number > not in the original list. > Changing the list is cheating. > Consider the following family of sets: > N_1 = {0} > N_2 = {0,1} > N_3 = {0,1,2} > ... > N_i = {0,1,..,i-1} > Earlier, someone (not me) said that the union of this > family of sets could be used as a definition of the > set of all natural numbers. That was me. > (If this was a valid definition of the set of all natural numbers, > we would not need the Axiom of Infinity.) How so? > Let K_i be the union of: N_1 u N_2 u ... u N_i. > Clearly, K_i is equal to N_i and this is true for all i. > If the set of all natural numbers, N, equals the > union of all N_i, then N must be equal to some N_i. This assumes that N is a finite set. It certainly does not prove it, though. > We can modify Cantor's proof to get a similar result. > For each N_i define a set M_i that contains every > element of N_i and the next ordinal. > N_1 = {0} M_1 = {0,1} > N_2 = {0,1} M_2 = {0,1,2} > ... > For all i, the union of M_1 u M_2 u ... u M_i > contains an element not in the union N_1 u N_2 u ... u N_i. > The union of all M_i contains an element > not in the union of all N_i. Again, this all works just fine if you are using only finite sets. It does not prove a thing about infinite sets. === Subject: Re: The list of all natural numbers don't exist >> I have provided two methods to find a number not in K. If K is supposed to be the set of all natural numbers, what sort of number is it that both is and is not in K? > I describe a method that creates a number not in the list > and you reply that, since the list is assumed to contain > every natural number, we can just add the number > the diagonal method created to our list and still have the same list. Either the diagonal method creates a natural number which is of necessity already in the list of all natural umbers, or it creates something which is not a natural number at all, or it fails to create anything at all. > By add 1 to this set > I assume you add a natural number that is not already in the set. > (If you mean add 1 to the largest natural in K, that's impossible, > since there is no largest natural in K.) > If there is a largest set of natural numbers there must exist > a largest natural number. Not so. The same argument would similarly require that the largest set of real numbers contain a largest real number and the largest set of rationals to contain a largest rational, but you are unable to name any of these supposedly largest members. > Since there is no largest natural number, there can be no > largest list of natural numbers Such a claim requires proof, of which you provide none. > Since K contains all the natural numbers, which one do we add > that got left out? Which natural number does your diagonal method > produce that is not in the set of all natural numbers? > Which real number does Cantor's proof create that > is not in the list of all real numbers? Everyone understands that the Cantor diagonal proof is just one, and not the primary one, of many proofs that there is no 'listing' of all reals. But neither that diagonal proof nor any of the other proofs of that theorem applies to listings of naturals. > The point of the proof is to show that there is at least > one natural number missing in any list of naturals numbers. > You can't just argue that your list contains all numbers > and the number created by the diagonal argument is > already in the list. I certainly can, and do, argue that all naturals are in the list. > The diagonal argument provably creates a number > not in the original list. > Changing the list is cheating. > Consider the following family of sets: > N_1 = {0} > N_2 = {0,1} > N_3 = {0,1,2} > ... > N_i = {0,1,..,i-1} > Earlier, someone (not me) said that the union of this > family of sets could be used as a definition of the > set of all natural numbers. > (If this was a valid definition of the set of all natural numbers, > we would not need the Axiom of Infinity.) > Let K_i be the union of: N_1 u N_2 u ... u N_i. > Clearly, K_i is equal to N_i and this is true for all i. > If the set of all natural numbers, N, equals the > union of all N_i, then N must be equal to some N_i. > We can modify Cantor's proof to get a similar result. > For each N_i define a set M_i that contains every > element of N_i and the next ordinal. > N_1 = {0} M_1 = {0,1} > N_2 = {0,1} M_2 = {0,1,2} > ... > For all i, the union of M_1 u M_2 u ... u M_i > contains an element not in the union N_1 u N_2 u ... u N_i. Irrelevant! > The union of all M_i contains an element > not in the union of all N_i. Wrong! The union of ALL M_j is precisely equal to the union of of ALL N_i since each is easily shown to be a subset of the other. This follows directly from such trivial facts as that for each i there is a j such that N_i is a subset of M_j, and for each j there is an i such that M_j is a subset of N_i. === Subject: Re: The list of all natural numbers don't exist If K is supposed to be the set of all natural numbers, what sort of > number is it that both is and is not in K? > ... > Everyone understands that the Cantor diagonal proof is just one, and > not the primary one, of many proofs that there is no 'listing' of all > reals. That's misleading. There is basically the diagonal proof, perhaps more correctly called the antidiagonal argument, and the powerset result, and the nested interval result. I find it more conscionable to interpret those differently than the status quo. consider for example the antidiagonal argument: with f(x) = x+1, the antidiagonal goes to the end of the list. With nested intervals, again having a notion of contiguous reals is a way to escape the consequences of inability of bijecting the naturals to the reals, _or_ well-ordering the reals in bijecting them to an ordinal. With the powerset result, in terms of the naturals, with the successor function, where there are ubiquitous ordinals, the unmapped element is the empty set, basically an escape route. So, 1) three is not many, and 2) it is not universally agreed what they mean, and 3) alternative theories can and do, as you note, explain applicable mathematics very well without transfinite cardinals. Please write out the list before replying. Ross === Subject: Re: The list of all natural numbers don't exist If K is supposed to be the set of all natural numbers, what sort of > number is it that both is and is not in K? > ... > Everyone understands that the Cantor diagonal proof is just one, and > not the primary one, of many proofs that there is no 'listing' of all > reals. > That's misleading. There is basically the diagonal proof, perhaps more > correctly called the antidiagonal argument, and the powerset result, > and the nested interval result. > I find it more conscionable to interpret those differently than the > status quo. consider for example the antidiagonal argument: with f(x) > = x+1, the antidiagonal goes to the end of the list. No it doesn't, because the list doesn't have an end. Or do you find it conscionable to simply consider the end of something that doesn't have an end, because lots of other things do have an end? > With nested > intervals, again having a notion of contiguous reals is a way to escape > the consequences of inability of bijecting the naturals to the reals, > _or_ well-ordering the reals in bijecting them to an ordinal. Indeed. I suppose once you're really practised at considering the end of something that doesn't have an end, considering a set of things such that between any two of them there is always a third, and considering that simultaneously there may be two of them next to each other, with no contradiction with the fact that between the adjactent two there is another one, comes as naturally as breathing. Brian Chandler http://imaginatorium.org === Subject: Re: The list of all natural numbers don't exist > My proof > You are very much confused about mathematical proof in general and set > theory in particular. Why don't you just read a book on mathematical > logic and one on set theory so that you don't continue wandering in a > fog about the subject while you so vigorously opine on it? I am more interested in proofs than set theory. If you have a problem with my proof, point out the problem. Don't just claim you have more knowledge of set theory. I'm sure you do know more set theory than I do, but so what? Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist If you have a problem with my proof, point out the problem. You proved that any finite enumeration of natural numbers is not an enumeration of all natural numbers. That's known to everybody. But it doesn't prove that the set of natural numbers does not exist; it doesn't prove that the set of natural numbers is not infinite (actually, it proves that the set of natural numbers is infinite); and it doesn't suggest any coherent application to refuting the proof that the set of real numbers is uncountable - whichever of these you might think is the case. > Don't just claim you have more knowledge of set theory. I didn't compare my personal knowledge with yours. I didn't mention anything about my own small core of knowledge. But if I were to compare, I'd say that my knowledge is not extensive, but your knowledge is inadequate for the purposes of the kind of things you wish to argue here. > I'm sure you do know more set theory than I do, but so what? It's not a matter of my meager knowledge being better than your nearly complete misunderstanding. It's a matter of your nearly complete misunderstanding (no matter with whom it is compared) preventing you from adressing the subject here coherently. So my positive suggestion is just to read a book on mathematical logic and one on set theory. MoeBlee === Subject: Re: The list of all natural numbers don't exist >> My proof > You are very much confused about mathematical proof in general and set > theory in particular. Why don't you just read a book on mathematical > logic and one on set theory so that you don't continue wandering in a > fog about the subject while you so vigorously opine on it? > I am more interested in proofs than set theory. > If you have a problem with my proof, point out the problem. > Don't just claim you have more knowledge of set theory. > I'm sure you do know more set theory than I do, but so what? That knowledge protects one from making an ass of oneself by claiming things that current set theories easily refutes. And if Russell is so interested in proofs, why is it that he ignores those of others and produces none of his own? === Subject: Re: The list of all natural numbers don't exist > But a list of natural numbers has a maximal element when it is finite. > And all lists of natural numbers are finite. > In that case there should be a largest list. What is that largest list? > Consider the list of lists of natural numbers. By your reasoning that > list must be finite. I am asking you to present the largest of all > those > lists. >> Just as there is no largest natural number, >> there is no largest list of natural numbers. >> If there was a largest list of natural numbers, >> it would be the list of all natural numbers. >> But, as this proof shows, there is no largest list. > Then, as a consequence of your proof, there cannot be a list of > all the natural numbers, because that list would be the largest > list. > Yes. Since there can be no largest list of natural numbers, > there can be no list of all natural numbers. > Suppose I have a set K, where > 1. 0 (a natural) is in K, > and > 2. if natural n is in K, then natural n+1 is also in K. > Your set theory tells us that K is a finite set. > My proof shows no such set exists. Your proof directly contradicts the Peano axioms which guarantee that such a set does exist. If your proof were valid, then every statement would have to be both true and false, and all arithmetic would be down the tubes.. > Since all > finite sets of naturals have a largest element (otherwise they > would not be finite), what is K_max, the largest element in K? > You tell me. > You are the one claiming to have the largest list of natural numbers. > What is the largest member of K? You are the only one who claims that there need be largest, so you are the only one responsible for proving its existence. We who claim that there is no such thing as a largest are quite happy not to have one. > Since your set theory says that there does not exist a list > (or set) of all the naturals, K cannot contain all the naturals. > So which naturals are not in K? The answer to this is trivial > if you can tell us what K_max is, since then K_max+1 could not > be in K. > I have provided two methods to find a number not in K. > One can use the diagonal argument to find such a number, No diagonal argument can prove the existence of what does not exist. > or one can take the union of all the natural numbers in K > and then add 1. That assumes falsely the the union of all natural numbers is a natural number. If one follows the NBG model, then n + 1 = n union {n}, so that the union of any *finite* set of naturals is the maximum of that set. So that anyone who claims that the set of all naturals is finite in that model is claiming that there is a maximum natural, which sensible people can see is false. === Subject: Re: The list of all natural numbers don't exist you keep talking about. > Every set of natural numbers has a largest member. > Proof: > Let: > {} = zero > {{}} = one > {{},{{}}} = two > ... > zero union one = {} u {{}} = {{}} > Given the members of a set, assume we can take the union > of the members of that set. > Given a set that contains only natural numbers, the union of the > members of that set equals the largest natural number in the set. Amazing. > Russell Kook. > - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist >> ... >> xxxx = four >> Are you claiming that four is not a natural number? >> No. I said that about the list of natural numbers, not about a >> finite >> sublist >> of the natural numbers. >> The formula I described works on any list of natural numbers. >> How does it work with an infinite list of natural numbers? > There is no such thing. The set of all natural numbers is just such a list. >> More strict, it is easily proven that the diagonal number has >> one more x than the last member of the list. This works well >> with lists that have a last element, and the proof applies only to >> lists that have a last member. The argument does not go through >> with lists that do *not* have a last element. >> It is more accurate to say the diagonal number has one more x >> than the largest number in the list. >> But a list of natural numbers has a maximal element when it is finite. > And all lists of natural numbers are finite. How is the list of all natural numbers finite? >> Of course, the argument above also shows that any list of natural >> numbers >> has a largest number. >> Not at all. > Where do you think the proof is flawed? > The proof is almost identical to the normal diagonal argument. >> I can provide other proofs in you want. >> I am not interested in that. Pray show how your argument shows that any >> list of natural numbers has a largest element. The only thing it shows >> is that any finite list of natural numbers has a largest element. We did >> know that already. > Perhaps you should first provide a definition of these infinite sets > you keep talking about. Countably infinite sets: The set of all whole numbers is an infinite set. The set of all natural numbers is an infinite set The set of all odd numbers is an infinite set The set of all even numbers is an infinite set The set of all prime numbers is an infinite set The set of all composite numbers is an infinite set The set of all rational numbers is an infinite set Uncountably infinite sets: The set of all real numbers between 0 and 1 is an infinte set (larger than these other sets) The set of all subsets of all sets > Every set of natural numbers has a largest member. > Proof: > Let: > {} = zero > {{}} = one > {{},{{}}} = two > ... > zero union one = {} u {{}} = {{}} > Given the members of a set, assume we can take the union > of the members of that set. > Given a set that contains only natural numbers, the union of the > members of that set equals the largest natural number in the set. What if the set has no largest number? > Russell > - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist > ... > xxxx = four > > Are you claiming that four is not a natural number? > > No. I said that about the list of natural numbers, not about a finite > sublist > of the natural numbers. > > The formula I described works on any list of natural numbers. > How does it work with an infinite list of natural numbers? > There is no such thing. There are lots of people for which there is such a thing. If you are not one of them, that is your loss, not theirs. > More strict, it is easily proven that the diagonal number has > one more x than the last member of the list. This works well > with lists that have a last element, and the proof applies only to > lists that have a last member. The argument does not go through > with lists that do *not* have a last element. > > It is more accurate to say the diagonal number has one more x > than the largest number in the list. > But a list of natural numbers has a maximal element when it is finite. > And all lists of natural numbers are finite. According to TO, maybe, but not to most people. > Of course, the argument above also shows that any list of natural > numbers > has a largest number. > Not at all. > Where do you think the proof is flawed? It assumes its conclusion. > The proof is almost identical to the normal diagonal argument. Almost doesn't cut it. > I can provide other proofs in you want. > I am not interested in that. Pray show how your argument shows that any > list of natural numbers has a largest element. The only thing it shows > is that any finite list of natural numbers has a largest element. We did > know that already. > Perhaps you should first provide a definition of these infinite sets > you keep talking about. The set of 'all' naturals, as defined in either ZFC or NBG. > Every set of natural numbers has a largest member. > Proof: > Let: > {} = zero > {{}} = one > {{},{{}}} = two > ... > zero union one = {} u {{}} = {{}} > Given the members of a set, assume we can take the union > of the members of that set. > Given a set that contains only natural numbers, the union of the > members of that set equals the largest natural number in the set. Non sequitur. You have not even shown yet that the union of two arbitrary naturals is a natural, much less that the union of an arbitrary family of naturals is a natural, and you need that, which is equivalent to your conclusion, in order to prove your conclusion. Which amounts to ASSUMING your conclusion in order to prove it. === Subject: Re: The list of all natural numbers don't exist >> Every set of natural numbers has a largest member. >> Proof: >> Let: >> {} = zero >> {{}} = one >> {{},{{}}} = two >> ... >> zero union one = {} u {{}} = {{}} >> Given the members of a set, assume we can take the union >> of the members of that set. >> Given a set that contains only natural numbers, the union of the >> members of that set equals the largest natural number in the set. > Non sequitur. > You have not even shown yet that the union of two arbitrary naturals is > a natural, much less that the union of an arbitrary family of naturals > is a natural, and you need that, which is equivalent to your conclusion, > in order to prove your conclusion. Are you claiming the union of two finite ordinals is not a finite ordinal? I am sure you can prove this much more elegantly than I can. Russell - 2 many 2 count === Subject: question about quaternions I have a basic question about quaternions: Say, we have an imaginary quaternion q = xi +yyj + zk , we can associate with it the 3x3 matrix R(q) = ( 0 ,-z, y ; z, 0 , -x ; -y, x, 0) . It is know always true that if q and r are imgainary quaternions, then qr is too, but as far as i understand (qr - rq)/2 does. So, we with each skewsymmetric real matrix 3x3 we can associate a quaternion q and the Lie algebra so(3,R) would be isomorphic with algebra of imaginary quaternions with the product not usual qr, but q x r = (qr - rq)/2. Am i right with this ? === Subject: Re: question about quaternions- >I have a basic question about quaternions: >Say, we have an imaginary quaternion q = xi +yyj + zk , we can >associate with it the 3x3 matrix >R(q) = ( 0 ,-z, y ; z, 0 , -x ; -y, x, 0) . It is know always true >that if q and r are imgainary quaternions, then qr is too, but as far >as i understand (qr - rq)/2 does. So, we with each skewsymmetric real >matrix 3x3 we can associate a quaternion q and the Lie algebra so(3,R) >would be isomorphic with algebra of imaginary quaternions with the >product not usual qr, but >q x r = (qr - rq)/2. Am i right with this ? Yes, you are right, as previous readers of your post already stated. Perhaps it is useful to recall the scalar-vector formula for quaternion multiplication in an old-fashioned apprearance but nevertheless invariant (*) form: Let q = (Sq, Vq) and r = (Sr, Vr) be quaternions in scalar-vector representation, then qr = (Sq.Sr - (Vq, Vr), Sq.Vr + Sr.Vq + Vq x Vr), or split into scalar and vector parts: S(qr) = Sq.Sr - (Vq, Vr), V(qr) = Sq.Vr + Sr.Vq + Vq x Vr. It is an instructive and not too easy exercise to give an invariant (*) proof of the skew-field property of the quaternions with the above definition of multiplication. The main difficulty is in the beginning: to give invariant definitions of the scalar and vector products of 3D vectors. A nice book on this subject is Antal E. Fekete: Real linear algebra. Marcel Dekker Inc. 1985, ISBN 0-8247-7238-5. Ciao: Johan E. Mebius (*) invariant: i.e. invariant under rotations of the 3D vector space of quaternion vector parts === Subject: Re: question about quaternions > I have a basic question about quaternions: > Say, we have an imaginary quaternion q = xi +yyj + zk , we can > associate with it the 3x3 matrix > R(q) = ( 0 ,-z, y ; z, 0 , -x ; -y, x, 0) . It is know always true > that if q and r are imgainary quaternions, then qr is too, but as far > as i understand (qr - rq)/2 does. So, we with each skewsymmetric real > matrix 3x3 we can associate a quaternion q and the Lie algebra so(3,R) > would be isomorphic with algebra of imaginary quaternions with the > product not usual qr, but > q x r = (qr - rq)/2. Am i right with this ? Your original statement about q and r being imaginary making qr imaginary is not true in general, but the outer product (qr - rq)/2 will always be. The inner product (qr + rq)/2 will always be real. === Subject: Re: question about quaternions > I have a basic question about quaternions: > Say, we have an imaginary quaternion q = xi +yyj + zk , we can > associate with it the 3x3 matrix > R(q) = ( 0 ,-z, y ; z, 0 , -x ; -y, x, 0) . It is know always true > that if q and r are imgainary quaternions, then qr is too, but as far > as i understand (qr - rq)/2 does. So, we with each skewsymmetric real > matrix 3x3 we can associate a quaternion q and the Lie algebra so(3,R) > would be isomorphic with algebra of imaginary quaternions with the > product not usual qr, but > q x r = (qr - rq)/2. Am i right with this ? If q and r are imaginary quaternions, then (qr - rq) / 2 is equal to cross product of vectors q and r, (qr - rq) / 2 = [q, r]. So you are right I think. === Subject: Re: SF: Does this work? > No, no, all this is good clean fun. I am even beginning to see what JSH > find in it You Are The One, Neo You Are Beginning To Believe === Subject: Re: SF: Easy math, theory behind the solution > That does not logically follow. > JSH is claiming that > (1) If RSA is broken, then the stock market will crash. > (2) The stock market is crashing. > Therefore > (3) RSA has been broken. > This is not valid reasoning; most people here will allow (1) and (2). > However, if P implies Q, and Q is true, then you CANNOT deduce that P > is also true. > --- Christopher Heckman Affirming the Consequent, logic 101. A well-known (and even named) logical fallacy. === Subject: Re: SF: Easy math, theory behind the solution <9CXig.86$7z4.37@fe06.lga That does not logically follow. > JSH is claiming that > (1) If RSA is broken, then the stock market will crash. > (2) The stock market is crashing. > Therefore > (3) RSA has been broken. > This is not valid reasoning; most people here will allow (1) and (2). > However, if P implies Q, and Q is true, then you CANNOT deduce that P > is also true. > Affirming the Consequent, logic 101. A well-known (and even named) > logical fallacy. --- Christopher Heckman === Subject: Re: SF: Easy math, theory behind the solution > He's lying. > Look at the stock market. > The world is treading water waiting on the result of this debate. > The answer is, I found an easy solution to the factoring problem. > Math people lie all the time. They're lying about my solution. > RSA is broken. > James Harris Oh, dear, I -knew- I should have bought milk and eggs at the grocery today! Grocery shelves will be as good as empty tomorrow because of what you have done! ... Or maybe not. === Subject: Re: SF: Easy math, theory behind the solution > Weird, in the version I know (which is exactly the same bar this > change), it's an Irishman. Do Cajun folk talk like Irish folk, then? The Irishman can keep yakking all he wants about math as long as he keeps his hands off my crawfish. === Subject: Cookbook for Ito calculus What do you recommend for a cookbook approach to stochastic calculus? I have the book by Bernt Oksendal, but I was looking for something more cookbookish, in the same style as books typically used in the USA for first semester engineering calculus (Thomas and Finney or Stewart) or tightly summarized like the Schaum Outline Series. === Subject: Re: Cookbook for Ito calculus > What do you recommend for a cookbook approach to stochastic calculus? I > have the book by Bernt Oksendal, but I was looking for something more > cookbookish, in the same style as books typically used in the USA for > first semester engineering calculus (Thomas and Finney or Stewart) or > tightly summarized like the Schaum Outline Series. Several financial engineering programs that I looked into use Steve Shreve's book (among those would be Carnegie Mellon and NYU IIRC). I do not know if it would suit your needs as I have not read it. I am just now taking calculus I with the intention of studying stochastic calculus a few semesters from now. So if you decide you like it, let us know - I'd be interested in your opinion of it. === Subject: THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a nano chip OR QUNATIUM COMPUTER BACK to adam and eve it would take them 3000 years to figure it out. Matrix Computing to Infinte Universes Infinte Outcomes know the all the Fundametal Laws of the Universe it would be a INSTANIOUS leap in technology INSTANIOUS TIME for INSTANIOUS TIME using A SIMPLE QUNATIUM COMPUTER thats mirrored against and its OWN NERUAL NETWORK AT THE LEVEL OF 40 -23 YOCTO THEN LET quantium AI THINK faster then the KNOWLEDGE there will be time dilation that the computer will now become 2 or inifnte INSTANIOULY with one ahead of itself to watch and learn form another simulationlie theone at alt.alien research with ininfte knowledge data base nerual network IN SIMULATION OPERATED ON THE FREQUENCES OF GOOD The knowledge of infinite beaty is 10 fold see compression www.speedypc.20m.com THEN LOOK AT QUNATATIVIE DEFINITIONS AND EVERY LAW CAN BE TRUE AND FALSE AT supporting everything LIKE the pyramind program and THE THEROY OF EVERYTHING, AND THAT is the quantium address of everything that is inside and outside our universe.By a magnetic field. Outside and inside of all that is contained there in.Finding the laws of all quantium physics and all-matter and everything that can be changed and from adjusting the frequency of that matter and space time, speed of light, magnetics, gravity ect. Searching for the one law of infinite dynamics. Picture the laws as a dotted circle ( pyramid program )and everything that can happen in that universe is because of knowing the All laws in that universe. REPLACE word INFINITE on the internet WITH CALCULABLE NOW INCALCULABLE IS NOW CALCUABLE!!! in ll forms of INCALCULABLE replace with INFINITE then with A.I now it's CALCUABLE! www.psitech.net/ run the internet in infinite order then over again with wordlist new defintions! with http://www.hyperdictionary.com/ === Subject: Re: THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a > THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a > nano chip OR QUNATIUM I have certainly learned a large number of new words by reading this post-- wonder what they mean? Here's my list..... QUNATIUM Infinte Fundametal INSTANIOUS NERUAL simultationle QUNATATIVE pyramind THEROY ect. defintions I also learned quite a large number of ways to mangle the English language. I guess it was in English. I suggest that the writer himself would do well to consult a dictionary...... Grover Hughes COMPUTER BACK to adam and eve it would take them > 3000 years to figure it out. Matrix Computing to Infinte Universes > Infinte Outcomes know the all the Fundametal Laws of the Universe it > would be a INSTANIOUS leap in technology INSTANIOUS TIME for > INSTANIOUS TIME using A SIMPLE QUNATIUM COMPUTER thats mirrored > against and its OWN NERUAL NETWORK AT THE LEVEL OF 40 -23 YOCTO THEN > LET quantium AI THINK faster then the KNOWLEDGE there will be time > dilation that the computer will now become 2 or inifnte INSTANIOULY > with one ahead of itself to watch and learn form another simulationlie > theone at alt.alien research with ininfte knowledge data base nerual > network IN SIMULATION OPERATED ON THE FREQUENCES OF GOOD The knowledge > of infinite beaty is 10 fold see compression www.speedypc.20m.com THEN > LOOK AT QUNATATIVIE DEFINITIONS AND EVERY LAW CAN BE TRUE AND FALSE AT > supporting everything LIKE the pyramind program and THE THEROY OF > EVERYTHING, AND THAT is the quantium address of everything that is > inside and outside our universe.By a magnetic field. Outside and > inside of all that is contained there in.Finding the laws of all > quantium physics and all-matter and everything that can be changed and > from adjusting the frequency of that matter and space time, speed of > light, magnetics, gravity ect. Searching for the one law of infinite > dynamics. Picture the laws as a dotted circle ( pyramid program )and > everything that can happen in that universe is because of knowing the > All laws in that universe. > REPLACE word INFINITE on the internet WITH CALCULABLE NOW INCALCULABLE > IS NOW CALCUABLE!!! in ll forms of INCALCULABLE replace with INFINITE > then with A.I now it's CALCUABLE! > www.psitech.net/ > run the internet in infinite order then over again with wordlist new > defintions! with http://www.hyperdictionary.com/ === Subject: Re: THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a > THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a > nano chip OR QUNATIUM COMPUTER BACK to adam and eve it would take them > 3000 years to figure it out. Matrix Computing to Infinte Universes > Infinte Outcomes know the all the Fundametal Laws of the Universe it > would be a INSTANIOUS leap in technology INSTANIOUS TIME for [... etc etc ...] Wow. I'll try one of whatever he's having. cordially Y.T. -- Remove YourClothes before you email me. === Subject: Re: THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a >> THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a >> nano chip OR QUNATIUM COMPUTER BACK to adam and eve it would take them >> 3000 years to figure it out. Matrix Computing to Infinte Universes >> Infinte Outcomes know the all the Fundametal Laws of the Universe it >> would be a INSTANIOUS leap in technology INSTANIOUS TIME for > [... etc etc ...] > Wow. > I'll try one of whatever he's having. he may not be comming back to this world. i'll pass. === Subject: Re: THE MATRIX IS PERFECT MORE COMPLEX THEN THE UNIVERSE and if you took a meth is bad === Subject: Re: Indra's Net of Pearls On Thu, 8 Jun 2006 01:37:19 +0300, Ioannis an infinite 3D grid of pearls which gets reflected on each ball. Not just >> three. >I have placed one more close-up of the central Indra Pearl here, with the >viewer positioned at (0.5,0.5,-0.5) >http://misc.virtualcomposer2000.com/indra.jpg (125 KB) Impressive as usual. Ioannis, several times you mentioned the Wada figure as using reflections in three balls. It is four! In a tetrahedral cannonball pile configuration. John === Subject: Re: Indra's Net of Pearls > On Thu, 8 Jun 2006 01:37:19 +0300, Ioannis an infinite 3D grid of pearls which gets reflected on each ball. Not just >> three. >I have placed one more close-up of the central Indra Pearl here, with the >viewer positioned at (0.5,0.5,-0.5) >http://misc.virtualcomposer2000.com/indra.jpg (125 KB) > Impressive as usual. > Ioannis, several times you mentioned the Wada figure as using > reflections in three balls. It is four! In a tetrahedral cannonball > pile configuration. And right you are, John. I kept staring at the central triangle in Wada's configuration that Brian gave the code for, trying to figure out how it generated all those patterns and finally it hit me, that there must be a ball there. Another curious thing with such configurations, if you exclude external objects added for visual effects, is the answer to the question what is it REALLY that gets reflected on those surfaces or that which we see there? A completely reflective sphere in void space is basically invisible in the absence of any external objects. In other words what the viewer sees on its surface, is only what gets reflected against it. But on its surface gets reflected other such spheres, for which the same assumption holds (i.e. they are invisible). This is true for EVERY reflection on EVERY sphere, ad infinitum, so if one excludes the images of the sphere boundaries which are shown because the rendering program forces some ambient light and some diffusion, it would appear that the image shown on any particular surface is basically a complex light pattern created by the only light source that exists on the scene. In other words, what we see is...infinitely many copies of the light source. This is quite puzzling to me. Assuming the light source is point-like/dimensionless, to me those patterns look like Cantor dust. > John -- Ioannis === Subject: Re: Indra's Net of Pearls >> On Thu, 8 Jun 2006 01:37:19 +0300, Ioannis > an infinite 3D grid of pearls which gets reflected on each ball. Not > just > three. >>I have placed one more close-up of the central Indra Pearl here, with the >>viewer positioned at (0.5,0.5,-0.5) >>http://misc.virtualcomposer2000.com/indra.jpg (125 KB) >> Impressive as usual. >> Ioannis, several times you mentioned the Wada figure as using >> reflections in three balls. It is four! In a tetrahedral cannonball >> pile configuration. > And right you are, John. I kept staring at the central triangle in Wada's > configuration that Brian gave the code for, trying to figure out how it > generated all those patterns and finally it hit me, that there must be a > ball there. > Another curious thing with such configurations, if you exclude external > objects added for visual effects, is the answer to the question what is it > REALLY that gets reflected on those surfaces or that which we see there? I have some Wada fractal images made with different raytracing programs webbed on http://rr.op.het.net/fract/wada.html -- -!- === Subject: Singular Value Decomposition Hi again. I'm still playing maths or better, learning the rules :) I'm always interested in linear algebra, and I find interesting the Singular Value Decomposition and its implications. So, any matrix can be decomposed in M = U S V^*, where ^* is the conjugate transpose, and in case M is real, then V^* = V^T, moreover, V and U are unitary, and as long as I understand, their transposed versions are their inverses. Am I right? What attracts me is S, the singular value. Can 0 belong to the singular values? If M is an m x n matrix, then as I read, S is the linear transformation K^n -> K^m so that if s is a singular value, T(v) = s u (with v in K^n and u in K^m), and being s = 0 would be weird to me... If M has full rank, are all the singular values different? What if M is diagonizable? If M is square, then are U, V and S related to the diagonalization? And last of all questions :) If a matrix A is diagonizable, then is S^T A S still diagonizable? I don't know, but S adds some 0 eigenvalues and adds dimension to the kernel of the resulting space. Is it true? -- Sensei The optimist thinks this is the best of all possible worlds. The pessimist fears it is true. [J. Robert Oppenheimer] === Subject: Re: Singular Value Decomposition > Hi again. I'm still playing maths or better, learning the rules :) > I'm always interested in linear algebra, and I find interesting the > Singular Value Decomposition and its implications. So, any matrix can > be decomposed in M = U S V^*, where ^* is the conjugate transpose, and > in case M is real, then V^* = V^T, moreover, V and U are unitary, and > as long as I understand, their transposed versions are their inverses. > Am I right? If M is real, then yes, their transposes are their inverses. Otherwise, you have to take a complex conjugate of all the terms as well. > What attracts me is S, the singular value. > Can 0 belong to the singular values? Yes. A zero singular value corresponds to a rank deficiency (i.e. your matrix is not full rank, and the number of ranks you're missing is the number of zero singular values) > If M is an m x n matrix, then as I read, S is the linear transformation > K^n -> K^m so that if s is a singular value, T(v) = s u (with v in K^n > and u in K^m), and being s = 0 would be weird to me... It just means you have a nontrivial null space, and there's not anything weird about that, is there? > If M has full rank, are all the singular values different? No, the identity matrix is full rank, and all its singular values are all one. > What if M is > diagonizable? If M is square, then are U, V and S related to the > diagonalization? Not necessarily, because the matrices that diagonalize M are generally not orthogonal. However if M is _symmetric_, then the spectral theorem tells you that U = V, and S contains the eigenvalues of M, so in this case they're certainly related. > And last of all questions :) If a matrix A is diagonizable, then is S^T > A S still diagonizable? I don't know, but S adds some 0 eigenvalues and > adds dimension to the kernel of the resulting space. Is it true? I'm not sure, it seems like that would require relating the eigenvectors of A and S^T A S in some way which is not obvious to me... === Subject: Re: Singular Value Decomposition >> What attracts me is S, the singular value. >> Can 0 belong to the singular values? >Yes. A zero singular value corresponds to a rank deficiency (i.e. your >matrix is not full rank, and the number of ranks you're missing is the >number of zero singular values) Conventions may differ, but it's quite common to define the singular values as the square roots of the _nonzero_ diagonal entries of S. See e.g. Strang, Linear Algebra and its Applications, Appendix A. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: JSH: So I lose I don't see any way to get through the complete social control that mathematician have in pure math areas, and I can't get a factoring solution, so there is nothing else for me to do in this area. The idea was for a treasure hunt. I'd look for simple proofs in dramatic areas and get rich if I could just fine one. But you can't get rich if no one will acknowledge your discovery, so an unacknowledged discovery, is no discovery at all. So by that brutal social logic, I have no discoveries. My analysis is that mathematicians will and do routinely lie about pure mathematical arguments, knowing that there is no other authority outside of their circles. My final analysis is that there is no way to get past this reality. Being someone who was in it for the money, there is no reason left to bang my head against a system that is so completely broken. At this time, it is impossible to determine what mathematicians say is true, and what is not true, as it's so convoluted and so involved with their own self-interest. I do hope that at least in applied mathematics, where lives can be lost dependent on whether or not the mathematics is true, these damn cons actually do tell the truth. I am now working to push myself away from the reality of having made what should have been major discoveries, with my short proof of Fermat's Last Theorem, my prime counting function, and my other various research finds, which turned out to not be because mathematicians would not accept them, to accepting the reality that we live in a stupid, senseless world. Yesterday I saw a news thing about some heads being cut off in Iraq. Not a big deal in today's world. Just another thing. This world is useless. There is nothing of value. There is nothing left for the discoverer. Humanity may as well be extinct. As it is so totally useless. It has no value. James Harris === Subject: Re: JSH: So I lose > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine [sic] one. [...] > Being someone who was in it for the money, there is no reason left to > bang my head against a system that is so completely broken. Hmm. I don't care about money or legitimacy in the public's eyes. -- James Harris, After all, I wouldn't give a damn about the prize money so I wouldn't need to even bother submitting them to RSA. -- James Harris, Can't you keep your story straight? > James Harris === Subject: Re: JSH: So I lose >>The idea was for a treasure hunt. I'd look for simple proofs in >>dramatic areas and get rich if I could just fine [sic] one. > [...] >>Being someone who was in it for the money, there is no reason left to >>bang my head against a system that is so completely broken. > Hmm. > I don't care about money or legitimacy in the public's eyes. > -- James Harris, > After all, I wouldn't give a damn about the prize money so I wouldn't > need to even bother submitting them to RSA. > -- James Harris, > Can't you keep your story straight? You'd have to be a _real_ genius to do that. === Subject: Re: JSH: So I lose I'm quoting this latest gem in full, unaltered and without comments, just in case someone wants to throw it back in your face after you've deleted it. > I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine one. > But you can't get rich if no one will acknowledge your discovery, so an > unacknowledged discovery, is no discovery at all. > So by that brutal social logic, I have no discoveries. > My analysis is that mathematicians will and do routinely lie about > pure mathematical arguments, knowing that there is no other authority > outside of their circles. > My final analysis is that there is no way to get past this reality. > Being someone who was in it for the money, there is no reason left to > bang my head against a system that is so completely broken. > At this time, it is impossible to determine what mathematicians say is > true, and what is not true, as it's so convoluted and so involved with > their own self-interest. > I do hope that at least in applied mathematics, where lives can be lost > dependent on whether or not the mathematics is true, these damn cons > actually do tell the truth. > I am now working to push myself away from the reality of having made > what should have been major discoveries, with my short proof of > Fermat's Last Theorem, my prime counting function, and my other various > research finds, which turned out to not be because mathematicians would > not accept them, to accepting the reality that we live in a stupid, > senseless world. > Yesterday I saw a news thing about some heads being cut off in Iraq. > Not a big deal in today's world. Just another thing. > This world is useless. There is nothing of value. There is nothing > left for the discoverer. > Humanity may as well be extinct. As it is so totally useless. > It has no value. > James Harris -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: So I lose > My analysis is that mathematicians will and do routinely lie about > pure mathematical arguments, knowing that there is no other authority > outside of their circles. ..... > At this time, it is impossible to determine what mathematicians say is > true, and what is not true, as it's so convoluted and so involved with > their own self-interest. If it is ...impossible to determine what mathematicians say is true, and what is not true... then how can you say that ...mathematicians will and do routinely lie about 'pure' mathematical arguments...? You would have to know that an argument that someone made is not true in order to say that he lied. Of course, as I reply to your troll bait, I know that you lie because I can determine the truth of your argument and what you have written you know not to be true. === Subject: Re: JSH: So I lose > I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. Why don't you add another couple of variables? === Subject: Re: JSH: So I lose <6this02dsf8w$.18emkkv3gg2oh.dlg@40tude.net> Bertie, This needs at least 3 more variables and even then he'd need to complete the square a few more times. Finally, he should rearrange the symbols and then back again several times. He needs to break the spirit of the evil algebra that is desperate to prevent him finding a solution. Square and rearrange, square and rearrange, eventually it will crack. This had better work - my life savings have just been wiped out by stock market crash. Why oh why does James go to print so early with his results? The markets are terrified. G. > I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. > Why don't you add another couple of variables? === Subject: Re: So I lose > Humanity may as well be extinct. As it is so totally useless. > It has no value. Is that a threat? === Subject: Re: JSH: So I lose [added JSH: to subject] [jstevh@msn.com] > I don't see any way to get through the complete social control that > mathematician have in pure math areas, They do insist on proofs. Your decade-long crusade has at least verified that strident attempts at bluff, bluster, bull, and bullying aren't > and I can't get a factoring solution, Not so far. > so there is nothing else for me to do in this area. > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine one. Do you really think you could, e.g., transform a short proof of FLT into Big Bux? Seems unlikely to me. Like, who would pay you for that, and why? The general public couldn't care less. If you had such a proof, I'm sure you could get it published in a technical journal, but if you think _that_ would make you rich you should do more investigation of potential payoffs next time around :-) > But you can't get rich if no one will acknowledge your discovery, so an > unacknowledged discovery, is no discovery at all. > So by that brutal social logic, I have no discoveries. You did come up with a correct recursive formulation of the prime-counting function. That's unique among your discoveries because it's provably correct. That's brutal mathematical logic, which is what you're really banging your head against. > My analysis is that mathematicians will and do routinely lie about > pure mathematical arguments, Then your analysis of that is wrong. > knowing that there is no other authority outside of their circles. > My final analysis is that there is no way to get past this reality. > Being someone who was in it for the money, there is no reason left to > bang my head against a system that is so completely broken. The system is doing what it's supposed to do here, maintaining high standards for proof which you simply haven't met. Worse, when mathematicians tell you flat-out that one of your purported proofs is not only wrong, but that the claim you're trying to prove is provably false, it's entirely your problem when you refuse to accept the truth. > At this time, it is impossible to determine what mathematicians say is > true, and what is not true, as it's so convoluted It may well be impossible for _you_ to tell. The cure for that is to learn some math. > and so involved with their own self-interest. LOL! Coming from someone who just said he was only in it for the money, you have scant claim to the moral high ground here. > I do hope that at least in applied mathematics, where lives can be lost > dependent on whether or not the mathematics is true, these damn cons > actually do tell the truth. Of course they do. > I am now working to push myself away from the reality of having made > what should have been major discoveries, with my short proof of > Fermat's Last Theorem, my prime counting function, and my other various > research finds, which turned out to not be because mathematicians would > not accept them, to accepting the reality that we live in a stupid, > senseless world. How self-centered are you? Don't answer that :-) > Yesterday I saw a news thing about some heads being cut off in Iraq. > Not a big deal in today's world. Just another thing. > This world is useless. There is nothing of value. There is nothing > left for the discoverer. Of course there is. Discoveries are made every day, quite possibly at a higher rate than at any previous time in history. > Humanity may as well be extinct. As it is so totally useless. > It has no value. Let's see. Because you don't know how to do proofs, humanity may as well be extinct. Ya, that's clear. > James Harris That wasn't in doubt :-)/:-( === Subject: Re: JSH: So I lose > I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. Dude! Just spend a few months working through Herstein. I'm telling you ... you do that, you'll have a whole new point of view of what it's all about. === Subject: Re: JSH: So I lose > Dude! Just spend a few months working through Herstein. You seem not to apprehend that he is unable to work through even the first few chapters of Herstein. He's mathematically challenged. > I'm telling you ... you do that, you'll have a > whole new point of view of what it's all about. And he's unwilling, because he is unable. === Subject: Re: So I lose >I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine one. > But you can't get rich if no one will acknowledge your discovery, so > an > unacknowledged discovery, is no discovery at all. > So by that brutal social logic, I have no discoveries. > My analysis is that mathematicians will and do routinely lie about > pure mathematical arguments, knowing that there is no other > authority > outside of their circles. > My final analysis is that there is no way to get past this reality. > Being someone who was in it for the money, there is no reason left > to > bang my head against a system that is so completely broken. > At this time, it is impossible to determine what mathematicians say > is > true, and what is not true, as it's so convoluted and so involved > with > their own self-interest. > I do hope that at least in applied mathematics, where lives can be > lost > dependent on whether or not the mathematics is true, these damn cons > actually do tell the truth. > I am now working to push myself away from the reality of having made > what should have been major discoveries, with my short proof of > Fermat's Last Theorem, my prime counting function, and my other > various > research finds, which turned out to not be because mathematicians > would > not accept them, to accepting the reality that we live in a stupid, > senseless world. > Yesterday I saw a news thing about some heads being cut off in Iraq. > Not a big deal in today's world. Just another thing. > This world is useless. There is nothing of value. There is nothing > left for the discoverer. > Humanity may as well be extinct. As it is so totally useless. > It has no value. Don't give up James!! Using your methods I got the tricky factorization 5579051284583946370062793502347650476748696836697060288885995066305155734471 0073205552063670462816641180583476458040210671833016029871169206301037423748 0 66113826045787271785770607780025833237812630915484433047698063073 = 5698767866324732345546790765456546865678765556765457655456456569765675448354 435545434544354555566644354389829 * 9789925498723649578236578926439782246392659273569236578265443597826439578436 958763275692165932856972346598637 I think you could crack RSA-704 !! -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: JSH: So I lose > I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine one. > But you can't get rich if no one will acknowledge your discovery, so an > unacknowledged discovery, is no discovery at all. RSA-704 $30,000 Not Factored RSA-768 $50,000 Not Factored RSA-896 $75,000 Not Factored RSA-1024 $100,000 Not Factored RSA-1536 $150,000 Not Factored RSA-2048 $200,000 Not Factored So you could have made at least 605,000 dollars, by factoring the above. Why didn't you? > So by that brutal social logic, I have no discoveries. Has nothing to do with social logic. You just had no discoveries, period. > This world is useless. There is nothing of value. There is nothing > left for the discoverer. Wrong again. > Humanity may as well be extinct. As it is so totally useless. Wrong again. > It has no value. Wrong again. > James Harris Wrong again. === Subject: Re: So I lose >I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. like most people. > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine one. get rich quick doesn't work. > But you can't get rich if no one will acknowledge your discovery, so an > unacknowledged discovery, is no discovery at all. you have not presented a discovery. Ideas yes, but no discovery > So by that brutal social logic, I have no discoveries. So you blame others for your perceived failures. > My analysis is that mathematicians will and do routinely lie about > pure mathematical arguments, knowing that there is no other authority > outside of their circles. lies. > My final analysis is that there is no way to get past this reality. you base your reality upon lies. > Being someone who was in it for the money, there is no reason left to > bang my head against a system that is so completely broken. Where is your discovery? > At this time, it is impossible to determine what mathematicians say is > true, and what is not true, as it's so convoluted and so involved with > their own self-interest. Of course they are beaming prozmatical energy into your brain to stop you from inventing new discoveries and getting rich. > I do hope that at least in applied mathematics, where lives can be lost > dependent on whether or not the mathematics is true, these damn cons > actually do tell the truth. Ask them and they will say We always lie > I am now working to push myself away from the reality of having made > what should have been major discoveries, with my short proof of > Fermat's Last Theorem, my prime counting function, and my other various > research finds, which turned out to not be because mathematicians would > not accept them, to accepting the reality that we live in a stupid, > senseless world. that is the reality you have chosen. > Yesterday I saw a news thing about some heads being cut off in Iraq. > Not a big deal in today's world. Just another thing. > This world is useless. There is nothing of value. There is nothing > left for the discoverer. > Humanity may as well be extinct. As it is so totally useless. > It has no value. Who is paying your bills? Do they think you have value to live? > James Harris === Subject: Re: So I lose >I don't see any way to get through the complete social control that > mathematician have in pure math areas, and I can't get a factoring > solution, so there is nothing else for me to do in this area. > The idea was for a treasure hunt. I'd look for simple proofs in > dramatic areas and get rich if I could just fine one. > But you can't get rich if no one will acknowledge your discovery, so an > unacknowledged discovery, is no discovery at all. > So by that brutal social logic, I have no discoveries. Aw, stop crying in your beer. For someone who is mathematically challenged, you have gotten a lot of mileage out of knowing very little. One reason for this is an unwarranted optimism on the part of the on-line mathematical community. With a few exceptions, they overestimated their ability to explain to you what is obvious to them. === Subject: S^n x |R parallelizable I don't get the proof for following problem: Show that S^n x |R is parallelizable for all n. Could you give me some hints? Lf === Subject: Re: S^n x |R parallelizable >I don't get the proof for following problem: >Show that S^n x |R is parallelizable for all n. >Could you give me some hints? Hint: an open subset of a parallelizable manifold is a parallelizable manifold. Lee Rudolph === Subject: Re: S^n x |R parallelizable Show that S^n x |R is parallelizable for all n. >Could you give me some hints? > Hint: an open subset of a parallelizable manifold is a parallelizable > manifold. > Lee Rudolph We didn't have this fact yet. Do you think of S^n x |R as an open subset of |R^(n+2)? I'll first have to proof you hint for that! I'll try that, but isn't there a more direct way? Lf === Subject: Re: S^n x |R parallelizable >>I don't get the proof for following problem: >>Show that S^n x |R is parallelizable for all n. >>Could you give me some hints? >> Hint: an open subset of a parallelizable manifold is a parallelizable >> manifold. >> Lee Rudolph >We didn't have this fact yet. What definition of parallelizable are you using? If, for example, your definition is an n-manifold M is parallelizable if there exist n continuous vectorfields X_1,...,X_n defined everywhere on M such that, for every point P of M, the n tangent vectors X_1(P),...,X_n(P) are a basis for the tangent space T_P(M), then the proof of my hint is ... very simple. There are other definitions, but I can't think of one for which the proof of my hint is any harder than it is for that one. >Do you think of S^n x |R as an open >subset of |R^(n+2)? I hope you mean n+1 not n+2. Anyway, it's not that I think of it that way, it's that it is *diffeomorphic* to such an open set, and--for (again) any definition of parallelizable that I can imagine--the property of being parallelizable is preserved by any diffeomorphism. In fact, I am giving you an illustration of a powerful technique for proving *many* theorems in differential topology: if you want to prove that a manifold M has a certain property, and you can show (1) that the property is invariant under diffeomorphism, while (2) M is diffeomorphic to a manifold that obviously has the property, then you're all set. >I'll first have to proof you hint for that! I'll try that, but isn't >there a more direct way? I can't imagine a more direct way, no. For instance, supposing that you are using the definition I gave earlier, you might say it would be more direct to write down the n vectorfields explicitly--but when you come to write down those n vectorfields, the most natural choice is going (I bet) to come via the embedding I've hinted at. Have you worked out the case n = 1 yet (ignoring the fact that S^1 itself is actually parallelizable)? How about the case n = 2? Get your hands dirty with those two examples, and I am sure everything else will work much more cleanly. Lee Rudolph === Subject: probability: inseparable events, multiple trials I'm trying to find a way to mathematically solve this problem. Let's say there is a grid, and every trial, a random square will be filled in. (We'll picture a 4x4 bingo board for example). If you can only win by creating a specific pattern (we'll say a straight line, 3 squares long to keep this simple), what are the odds that in N trials, you will be able to win with this pattern. Since the trials are inseparable, picking any one square will change the number of possible squares that will still lead to a win in N turns. Still seems fairly straightforward. Now, to make it more challenging, lets say that some of the squares on this board are blocked, and cannot be used. This means that the total number of possible solutions is reduced, and some squares may no longer be able to provide a solution. Is it possible to still solve this problem mathematically? Your thoughts, anyone? === Subject: Re: probability: inseparable events, multiple trials > Now, to make it more challenging, lets say that some of the squares on > this board are blocked, and cannot be used. This means that the total > number of possible solutions is reduced, and some squares may no longer > be able to provide a solution. > Is it possible to still solve this problem mathematically? Your > thoughts, anyone? Yes. State the exact problem, and it will be solved. === Subject: Re: probability: inseparable events, multiple trials <10%ig.168751$Fs1.18010@bgtnsc05-news.ops.worldnet.att.net> Most of it's right there, sorry if it was confusing. I'll try to phrase it better. You have a bingo board with dimensions H (height) and W (width), leaving of course (H.87W) total squares. There are some number (N), greater than (H.87W), different balls that can be chosen. Any number of squares on the given board may be either free spaces, or unusable. We'll use a hook pattern as the only solution for now (identical to a knight's movement in chess): (the x's are selected) x x x x How would you find the odd's of creating this pattern in the given board when selecting X different random spaces? === Subject: Re: probability: inseparable events, multiple trials <10%ig.168751$Fs1.18010@bgtnsc05-news.ops.worldnet.att.net> Not sure.. Seems tougher to do because of the blocked spaces... === Subject: Need help with the probability question? A side effect of a certain anesthetic used in surgery is the hiccups, which occurs in about ten percent of cases. If three patients are scheduled for surgery today and are administered anesthetic, compute the probability that: 3 get the hiccups: 0 get the hiccups: at least one: 2 get the hiccups: one gets the hiccups: Please help me figure this out. === Subject: Re: Need help with the probability question? <22347960.1150058883444.JavaMail.jakarta@nitrogen.mathforum.org A side effect of a certain anesthetic used in surgery is the hiccups, > which occurs in about ten percent of cases. If three patients are > scheduled for surgery today and are administered anesthetic, compute the > probability that: > 3 get the hiccups: 10^-3 > 0 get the hiccups: 9^-3 > at least one: 1 - 9^-3 > 2 get the hiccups: #1 & 2 get hiccups and 3 doesn't + #2 & 3 get hiccups and 1 doesn't + #3 & 1 get hiccups and 2 doesn't > one gets the hiccups: #1 get hiccups and 2 & 3 don't + #2 get hiccups and 3 & 1 don't + #3 get hiccups and 1 & 2 don't === Subject: Re: Need help with the probability question? <22347960.1150058883444.JavaMail.jakarta@nitrogen.mathforum.org> which occurs in about ten percent of cases. If three patients are > scheduled for surgery today and are administered anesthetic, compute the > probability that: > 3 get the hiccups: > 10^-3 > 0 get the hiccups: > 9^-3 Wrong. > at least one: > 1 - 9^-3 Right idea, but wrong number. > 2 get the hiccups: > #1 & 2 get hiccups and 3 doesn't > #2 & 3 get hiccups and 1 doesn't > #3 & 1 get hiccups and 2 doesn't > one gets the hiccups: > #1 get hiccups and 2 & 3 don't > #2 get hiccups and 3 & 1 don't > #3 get hiccups and 1 & 2 don't Right. --- Christopher Heckman === Subject: uniform convergence I want to show that the sequence of functions { cos(x+x^2/(n*Pi))} is not uniform convergent to cos(x) in all R. I let x=n*Pi and thus cos(n*Pi+(n*Pi)^2/(n*Pi)) =cos(2*n*Pi) converges to 1. but cos(n*Pi) does not converge. is this enough argument? === Subject: Re: uniform convergence > I want to show that the sequence of functions { cos(x+x^2/(n*Pi))} is not > uniform convergent to cos(x) in all R. > I let x=n*Pi and thus cos(n*Pi+(n*Pi)^2/(n*Pi)) =cos(2*n*Pi) converges to 1. > but cos(n*Pi) does not converge. > is this enough argument? I would say, no. I would say, you need to use the definition of uniform convergence. === Subject: Japanese Temple Geometry I am interested in a book called Japanese Temple Geometry Problems Sangaku by Fukagawa. Anyone know where I could get a copy? Amazon doesn't have it. Recommendations for anything similar are also welcome. === Subject: Re: Japanese Temple Geometry > I am interested in a book called Japanese Temple Geometry Problems > Sangaku by Fukagawa. Anyone know where I could get a copy? Amazon > doesn't have it. Recommendations for anything similar are also welcome. I believe the book is available from The Charles Babbage Research Centre. Their contact info is at http://www.paddle.mb.ca/ArsComb/ArsComb.html === Subject: Re: Japanese Temple Geometry >I am interested in a book called Japanese Temple Geometry Problems > Sangaku by Fukagawa. Anyone know where I could get a copy? Amazon > doesn't have it. Recommendations for anything similar are also welcome. http://www.amazon.com/gp/product/0919611214/104-9668553-6096712?v=glance&n=2 83155 === Subject: Re: Japanese Temple Geometry <448d1c28$0$2015$9a622dc7@news.kpnplanet.nl> That's sort of what I meant by Amazon doesn't have it, b/c you can't order it fromt them >I am interested in a book called Japanese Temple Geometry Problems > Sangaku by Fukagawa. Anyone know where I could get a copy? Amazon > doesn't have it. Recommendations for anything similar are also welcome. > http://www.amazon.com/gp/product/0919611214/104-9668553-6096712?v=glance&n=2 8 3155 === Subject: Re: Proofs by induction and infinity Ross Clement (Email address invalid - do not use) > I note that various web pages talking about proof by induction talk > about n going up to infinity.... You've had several replies about that sloppy language. > The reason I ask is that I've noted that the standard proofs of basics > such as addition being commutative seem to be proved for whole numbers. > How would you then prove that addition is commutative on the real > numbers (including irrational ones).... If you're interested in building up the other standard number systems from the natural numbers, a standard reference is the small but rigorous book by Edmund Landau, Foundations of Analysis, translated from the easy-to-read German Grundlagen der Analysis. Ken Pledger. === Subject: Re: Proofs by induction and infinity If you're interested in building up the other standard number > systems from the natural numbers, a standard reference is the small but > rigorous book by Edmund Landau, Foundations of Analysis, translated > from the easy-to-read German Grundlagen der Analysis. At present I'm reading Dantzig's book on the history of maths Number: The Language of Science. As I walk to and from work, which takes 40 minutes each way, that's my primary reading time for background material (relative to my job). I'll look into this book as a successor to Dantzig. At present I'm finding that Dantzig is a good interesting introduction to some basic concepts, and I'm then hoping to refine those with future reading before getting back to books more immediately relevant to my job. I'm sure I covered many of these concepts when I was in school, but have strong reason to believe that revision will be very useful before trying to go forward again. There's a Japanese saying don't build a house on sand. In the meantime, where can I find a proof that the number of irrational numbers greatly exceeds the number of rational numbers? Ross-c === Subject: Re: Proofs by induction and infinity > In practice you rarely need transfinite induction for anything. Not true. Transfinite induction (or related methods, e.g. maximal principles such as Zorn's Lemma) are used often in contemporary mathematics. Without such capabilities one would essentially be restricted to finite mathematics, which'd be woefully impotent and painfully incomplete, e.g. --Bill Dubuque === Subject: Need an an example of a function I'm looking for a specific example of a function f: R^2 -> R that has partial derivatives of all orders, but is not continuous at the origin. === Subject: Re: Need an an example of a function >I'm looking for a specific example of a function f: R^2 -> R that has >partial derivatives of all orders, but is not continuous at the origin. Let g(u) = exp(-1/u^2), u ~=0, and 0 if u = 0. Define f(x, y) = g(x)*g(y)/(g(x)^2 + g(y^2)) except at the origin, and 0 at the origin. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Need an an example of a function reply-type=response NotP nous a r.8ecemment amicalement signifi.8e : > I'm looking for a specific example of a function f: R^2 -> R that has > partial derivatives of all orders, but is not continuous at the > origin. for x and y = 0 f(x,y) = 0 otherwise f(x,y) = xy/(x^2 + y^2) for y = 0, f(x,y) has partial derivatives in x of any order for y != 0, f(x,y) has also partial derivatives in x of any order and idem for partial derivatives in y. lim(x->0, f(x,x)) = 1/2 != f(0,0) : f is not continuous at (0,0). -- Patrick === Subject: Re: Need an an example of a function >NotP nous a r.8ecemment amicalement signifi.8e : >> I'm looking for a specific example of a function f: R^2 -> R that has >> partial derivatives of all orders, but is not continuous at the >> origin. >for x and y = 0 f(x,y) = 0 > otherwise f(x,y) = xy/(x^2 + y^2) >for y = 0, f(x,y) has partial derivatives in x of any order >for y != 0, f(x,y) has also partial derivatives in x of any order >and idem for partial derivatives in y. >lim(x->0, f(x,x)) = 1/2 != f(0,0) : f is not continuous at (0,0). But the mixed partial derivative f_{1,1}(x,0) = -1/x^2 -> 0 as x -> 0, so f_{2,1}(0,0) doesn't exist. Thus you don't have mixed partial derivatives of all orders. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Need an an example of a function Robert Israel nous a r.8ecemment amicalement signifi.8e : > I'm looking for a specific example of a function f: R^2 -> R that > has partial derivatives of all orders, but is not continuous at the > origin. >> for x and y = 0 f(x,y) = 0 >> otherwise f(x,y) = xy/(x^2 + y^2) >> for y = 0, f(x,y) has partial derivatives in x of any order >> for y != 0, f(x,y) has also partial derivatives in x of any order >> and idem for partial derivatives in y. >> lim(x->0, f(x,x)) = 1/2 != f(0,0) : f is not continuous at (0,0). > But the mixed partial derivative f_{1,1}(x,0) = -1/x^2 -> 0 as x -> 0, > so f_{2,1}(0,0) doesn't exist. Thus you don't have mixed partial > derivatives of all orders. Oups! You're right I just looked at homogeneous partial derivative ! -- Patrick, sorry === Subject: Re: Need an an example of a function > I'm looking for a specific example of a function f: R^2 -> R that has > partial derivatives of all orders, but is not continuous at the origin. Play around with the complex function exp(-1/z^4) for inspiration. === Subject: Re: Need an an example of a function > I'm looking for a specific example of a function f: R^2 -> R that has > partial derivatives of all orders, but is not continuous at the origin. Stan Wagon gives this example: [Discontinuous, But Partial Derivatives Exist] http://stanwagon.com/wagon/mathimages/Links/MathImages_lnk_7.html f(0,0) = 0 f(x,y) = (1/4) [1 + (x^2 + y^2)^(3/2)] cos(4 arctan(y, x)) otherwise Of course the partial derivatives are also not continous at the origin. === Subject: Re: Need an an example of a function >> I'm looking for a specific example of a function f: R^2 -> R that has >> partial derivatives of all orders, but is not continuous at the origin. > Stan Wagon gives this example: > [Discontinuous, But Partial Derivatives Exist] > http://stanwagon.com/wagon/mathimages/Links/MathImages_lnk_7.html > f(0,0) = 0 > f(x,y) = (1/4) [1 + (x^2 + y^2)^(3/2)] cos(4 arctan(y, x)) otherwise > Of course the partial derivatives are also not continous at the origin. But what does arctan(y,x) mean? (I'm not used to arctan being a function of two arguments) === Subject: Re: Need an an example of a function I'm looking for a specific example of a function f: R^2 -> R that has >> partial derivatives of all orders, but is not continuous at the origin. > Stan Wagon gives this example: > [Discontinuous, But Partial Derivatives Exist] > http://stanwagon.com/wagon/mathimages/Links/MathImages_lnk_7.html > f(0,0) = 0 > f(x,y) = (1/4) [1 + (x^2 + y^2)^(3/2)] cos(4 arctan(y, x)) otherwise > Of course the partial derivatives are also not continous at the origin. > But what does arctan(y,x) mean? (I'm not used to arctan being a function of > two arguments) Normally we would say arctan(y/x) to denote the angle the ray from the origin to (x,y) makes with the positive x-axis, but this only works for the first and fourth quadrants, because of how the period of tangent has length pi. Extending this function to two arguments lets us cover all four quadrants (which is not a big deal, given that we are multiplying in this case by 4, so the distinction between pi and 2*pi) doesn't matter), and it makes the definition clear when x = 0, ie. at the outside boundaries of the first and fourth quadrant. To sum up, it means the angle between the positive x-axis and the ray through the point (x,y). It's the same as arg(x+iy) in connection with complex numbers. Multiplying by 4 makes the cosine of the result cycle through a complete period in each quadrant. === Subject: Re: Need an an example of a function > I'm looking for a specific example of a function f: R^2 -> R that has > partial derivatives of all orders, but is not continuous at the origin. My apologies. I don't know whether such a function exists or not. === Subject: Re: Need an an example of a function >> I'm looking for a specific example of a function f: R^2 -> R that has >> partial derivatives of all orders, but is not continuous at the origin. > My apologies. I don't know whether such a function exists or not. I'm pretty sure such a function exists, but I haven't been able to find one. === Subject: Re: Need an an example of a function >> I'm looking for a specific example of a function f: R^2 -> R that has >> partial derivatives of all orders, but is not continuous at the origin. > My apologies. I don't know whether such a function exists or not. > I'm pretty sure such a function exists, but I haven't been able to find one. Why do you expect to be able to find a function which has a derivative at a point at which it is not continuous? === Subject: Re: Need an an example of a function > I'm looking for a specific example of a function f: R^2 -> R that has > partial derivatives of all orders, but is not continuous at the > origin. > My apologies. I don't know whether such a function exists or not. >> I'm pretty sure such a function exists, but I haven't been able to find >> one. > Why do you expect to be able to find a function which has a derivative > at a point at which it is not continuous? Well, I said partial derivatives. Existence of partial derivatives does not imply continuity. === Subject: Re: Need an an example of a function > I'm looking for a specific example of a function f: R^2 -> R that has >> partial derivatives of all orders, but is not continuous at the >> origin. >> My apologies. I don't know whether such a function exists or not. >> I'm pretty sure such a function exists, but I haven't been able to find > one. >> Why do you expect to be able to find a function which has a derivative >> at a point at which it is not continuous? > Well, I said partial derivatives. Existence of partial derivatives does > not imply continuity. Partial to what? === Subject: Re: Need an an example of a function > I'm looking for a specific example of a function f: R^2 -> R that has > partial derivatives of all orders, but is not continuous at the origin. No such function exists. Did you perhaps mean not differentiable at the origin? === Subject: Integer solutions I would greatly appreciate any comments upon the correctness of the following statement. Statement: The following equation has no integer solutions under the given conditions. (n^1/2 + m^1/2)^1/2 + (n^1/2 - m^1/2)^1/2 = 2^1/2(n^1/2 + c^1/2)^1/2 (1) Conditions: m, n, c are relatively prime integers, each > 1, none is a perfect square and m is even. My Proof: (1) can be written as (2) x^1/2 + y^1/2 = (2z )^1/2 (2) where x = (n^1/2 + m ^1/2), y = (n^1/2 - m^1/2) , z = (n^1/2 + c^1/2) x + y + 2(xy)^1/2 = 2z (3) (3) can be satisfied only if x + y = 2z (4) and xy = 0 (5) Since x can not be zero one takes y = 0 which implies m = n. But this is a contradiction because m, n are relatively prime integers each > 1. This justifies the statement. === Subject: Re: Integer solutions > I would greatly appreciate any comments upon the correctness of the > following statement. Your use of notation is sloppy. What's 2^1/2n, for example? > Statement: The following equation has no integer solutions under the > given conditions. > (n^1/2 + m^1/2)^1/2 + (n^1/2 - m^1/2)^1/2 = 2^1/2(n^1/2 + c^1/2)^1/2 > (1) 2.sqr n + 2.sqr(n - m) = 2.sqr n + 2.sqr c n = 5, m = 2, c = 3 > Conditions: m, n, c are relatively prime integers, each > 1, none is a > perfect square and m is even. Conditions exceeded. > My Proof: (1) can be written as (2) > x^1/2 + y^1/2 = (2z )^1/2 (2) > where x = (n^1/2 + m ^1/2), y = (n^1/2 - m^1/2) , z = (n^1/2 + c^1/2) > x + y + 2(xy)^1/2 = 2z (3) > (3) can be satisfied only if x + y = 2z (4) and xy = 0 (5) Notice, x and y are not integers. x = 1 = y and z = 2 is an instant of a family of solutions y = 1/x z = x/2 + 1/2x + 1 > Since x can not be zero one takes y = 0 which implies m = n. But this > is a contradiction because m, n are relatively prime integers each > 1. > This justifies the statement. === Subject: Re: Killer Sudoku <447363C7.2040005@netscape.net> <4av872107s65hd1eksabqi65g45r01skjn@4ax.com> I found a neat vriant in a new free tabloid with a business orientation. it's the 9x9 with 3x3 submagicsquares, with the addition of 4 *more* 3x3s, placed so that there's a border of empty boxes all around them; makes it sort of 4-dimensional, and also quicker to solve. I also saw one in *Discover*, that was 5x5 with rather arbitrary subgroups of five boxes (just connected by edges). thus: I'm not sure, and hydrogen is addressed what its relative energy-density is. the question is, how does it compare to methane, from which 99.44% of all H2 is derived? >Does this mean more ethanol is required per stroke? Is there an available >comparison with gaseous hydrogen? thus: free your mind from Newtonian slavery; thank you! >just what would be an axample of an effect, >if the relative speed of the medium were a consideration --it takes some to jitterbug! http://members.tripod.com/~american_almanac http://larouchepub.com/other/2006/3322_ethanol_no_science.html === Subject: Re: Killer Sudoku <447363C7.2040005@netscape.net> <4av872107s65hd1eksabqi65g45r01skjn@4ax.com> it's so clear, seeing the grey- hatched 3x3 submagicsquares, over the white ones; I meant, there's a border of the smallest boxes, that take a number, around the four extra 3x3s meaning also just one box-width between'em. the ones in the 5x5 sudoku, aside from the columns & rows, are worms e.g., but not exclusively. I'm sure, you can find a tonnage of this stuff on the highfalutin'net. >it's the 9x9 with 3x3 submagicsquares, >with the addition of 4 *more* 3x3s, placed so that >there's a border of empty boxes all around them; >makes it sort of 4-dimensional, and also quicker to solve. >I also saw one in *Discover*, >that was 5x5 with rather arbitrary subgroups >of five boxes (just connected by edges). --it takes some to jitterbug! http://members.tripod.com/~american_almanac http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html http://larouchepub.com/other/2006/3322_ethanol_no_science.html http://www.wlym.com/pdf/iclc/howthenation.pdf === Subject: Arranging m objects into n piles I have m objects which I want to put into n piles. Each pile has to have atleast one object and no two piles can have the same number of objects. I want to know the number of such arrangements which we can denote asP(m,n). For example, if m = 14 (objects) and n = 4 (piles) then P(14,4) = 5 and the arrangements are: 1 2 3 8 1 2 4 7 1 2 5 6 1 3 4 6 2 3 4 5 It is easy to write a program to compute these arrangements. For example, P(50,8) = 116. I was wondering if there is a formula or difference equation for P(m,n). === Subject: Re: Arranging m objects into n piles > I have m objects which I want to put into n piles. > Each pile has to have atleast one object and no two > piles can have the same number of objects. I want to > know the number of such arrangements which we can > denote as P(m,n). For example, if m = 14 (objects) and > n = 4 (piles) then P(14,4) = 5 and the arrangements are: > 1 2 3 8 > 1 2 4 7 > 1 2 5 6 > 1 3 4 6 > 2 3 4 5 > It is easy to write a program to compute these > arrangements. For example, P(50,8) = 116. I was > wondering if there is a formula or difference > equation for P(m,n). I believe in combinatorics your P(m, n) would be the number of distinct partitions of m into *exactly* n parts. For example, the distinct partitions of 8 are 8 7 1 6 2 5 3 5 2 1 4 3 1 ..so there are 6 distinct partitions of 8, usually denoted as q(8) = 6. With your notation we see that P(8, 1) = 1, P(8, 2) = 3, and P(8, 3) = 2. In general, the number of distinct partitions can be described as the coefficients of the generating function prod_{k = 1}^{oo} (1 + x^k) = (1 + x)(1 + x^2)... = 1 + x + x^2 + 2x^3 + 2x^4 + 3x^5 + 4x^6 + 5x^7 + ... = sum_{n = 0}^{oo} q(n) x^n. In the infinite product above we are choosing either 1 or x^i from every bracket and so we arrive at the complete number of distinct partitions. I'm not very good at combinatorics, but I'd assume that a similar generating function along these lines could be deduced for your P(m, n). I did a quick Google search and found http://www.mathpages.com/home/kmath556/kmath556.htm ..which you may find useful; hope that helps a bit. Kyle Czarnecki === Subject: Re: Arranging m objects into n piles >I have m objects which I want to put into n piles. Each pile >has to have atleast one object and no two piles can have >the same number of objects. I want to know the number of >such arrangements which we can denote asP(m,n). ... they contain 1, in which case taking away 1 from each gives a partition of n-m with m-1 parts (discounting the zero), or they do not, in which case this reduction gives a partition of n-m with exactly m parts. More importantly, this operation is bijective, making the recurrence obvious. (Thinking in terms of Ferrer's diagrams might make this more clear.) >It is easy to write a program to compute these >arrangements. For example, P(50,8) = 116. I was >wondering if there is a formula or difference equation >for P(m,n). Certainly: P(n,0) = { 1, n = 0; 0, n != 0 }, P(n,m) = P(n-m,m) + P(n-m,m-1). --- Stan Liou === Subject: SF: Square roots were the key? For years I've been looking for a way to factor one number using the factorization of some different number, which I call the surrogate for a method I call surrogate factoring or SF. Trouble is, every set of equations I played with would either factor pathetically, as in with a very low success rate, or I'd find that I'd just go in a circle needing the factorization of my target. As a very simple example that isn't exactly what I used consider something like T = (3x - 2y)(x+2y) so if you solve it out for a factorization you have 3x - 2y = f_1, and x + 2y = f_2 well, of course, if you solve the thing, some of the ways I was doing, where'd I'd do completions of the square, you just get what you'd have if you solve out the two linear equations. Well I pondered that for a while, and wonder, what if I give the algebra an ambiguous expression that can't have a single solution? So after a bit I came up with S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) where S is the surrogate, so that the math does NOT see the target T at all, for this first step, so it can't make everything rely on the factorization of T, but the expression is ambiguous because of the square roots. Yet, of course, it's trivial to find squares that will fit in for some S that you pick, like S=15. So what do you do with the thing? You multiply out, and get rid of the radicals by squaring to get handle terms with sqrt(xy), and then you complete the square twice. Someone posted a result of that on sci.math where that person used math software which is a good idea as it's a mess. Next you focus on the term that now doesn't have x or y, and based on what was posted on sci.math, it will have S^2 as a factor, and a lot of complicated stuff with k_1, k_2, k_3, and k_4, and you take that complicated stuff and set it to T, your target composite to factor. If that person's result was correct, then if you can use math software, I'd recommend it. Next part is where I don't know as you have k's raised to the 4, but you have 4 degrees of freedom, meaning you can set at least two of the k's to just about anything, other than 0, and then figure out what value for the next to last, will give you an integer for the last one. If you can manage that--that's the hard part, where if it can't be done practically, then this is yet another pure math curiousity--then you just go back to S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) and you can even use, like, S=15, and then you solve for x and y, like k_1*sqrt(x) + k_2*sqrt(y) = 5 k_3*sqrt(x) + k_4*sqrt(y) = 3 and with x and y, you can I think just switch signs there, with the factors of your target popping out, or you can go back to that messy expression and plug in your values for x and y to get back a factorization of your target. The target gets incidentally factored. If it can be made to work, it looks like it'll take a lot of expertise far outside what I usually do, as I focus on rather simple algebra most of the time. The hardest thing being getting integer k's out of that monster expression where the highest exponent is 4. Though I guess even that may require factoring T, so the algebra could still force you to know the factorization of T anyway, which I suspect is what happens, as that is what usually shoots down my surrogate factoring ideas. They kind of just go in a big freaking circle what I call a BFC, where you need the factorization of T anyway. James Harris === Subject: Re: Square roots were the key? > * TROLL !! * <<< ...for years I have been.... trying to master high school algebra === Subject: Re: Square roots were the key? > > * TROLL !! * <<< > ...for years I have been.... wasting my life with useless crap === Subject: Re: Square roots were the key? | | > > * TROLL !! * <<< | > ...for years I have been.... | wasting my life with useless crap Very depressing. OCD for trolls. Who can help him? -- Ed. ----------------------------------------------------- hex->bin->b64 F9E7707A2AF502D0A899C6ACB43A2D35EB7E === Subject: Re: Square roots were the key? > For years I've been looking for a way to factor one number using the > factorization of some different number, which I call the surrogate > for > a method I call surrogate factoring or SF. > Trouble is, every set of equations I played with would either factor > pathetically, as in with a very low success rate, or I'd find that > I'd > just go in a circle needing the factorization of my target. > As a very simple example that isn't exactly what I used consider > something like > T = (3x - 2y)(x+2y) > so if you solve it out for a factorization you have > 3x - 2y = f_1, and x + 2y = f_2 > well, of course, if you solve the thing, some of the ways I was > doing, > where'd I'd do completions of the square, you just get what you'd > have > if you solve out the two linear equations. > Well I pondered that for a while, and wonder, what if I give the > algebra an ambiguous expression that can't have a single solution? > So after a bit I came up with > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > where S is the surrogate, so that the math does NOT see the target T > at > all, for this first step, so it can't make everything rely on the > factorization of T, but the expression is ambiguous because of the > square roots. Excellent!! But why not use T = arcsin(3x - 2y+k_3+f_6+w) * arccos(x+2y+4k_9-k_17+V_max) ? These inverse trig varmints are even ambiguouser than sqrt. Way to go!! -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Square roots were the key? > Excellent!! > But why not use > T = arcsin(3x - 2y+k_3+f_6+w) * arccos(x+2y+4k_9-k_17+V_max) > These inverse trig varmints are even ambiguouser than sqrt. Way to > go!! LOL === Subject: Re: Square roots were the key? > As a very simple example that isn't exactly what I used consider > something like > T = (3x - 2y)(x+2y) Seriously, let's take a closer look at this. For the number you want to factor, T, it is a multiple of two prime numbers, let's call them X and Y. For the moment, you really don't need all the other information there. So, upon closer inspection, we can simplify your formula: T = X . Y That's the -first- thing you can do to make your life easier. You can even plug it into your graphing calculator. Let T=15, and let the graphing function read Y=15/X. When X=3, Y=5. When X=5, Y=3. See? Easy. Whole numbers are good. BTW, nice -hyperbola-. > so if you solve it out for a factorization you have... 3=X (or f_1, as you call it) and 5=Y (or f_2) and 15=T. > So after a bit I came up with > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) Above I made your life simpler, now you are making it even more complex, unnecessarily. Mathematics likes simplicity. Let's take a closer look at Proginoske's post, again, where he simplified this formula for us: S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) S = (k_1 k_3 x + k_2 k_4 y) + (k_2 k_3 + k_1 k_4) sqrt(xy) The modified S is: S=(k_1 k_3 x + k_2 k_4 y) - (k_2 k_3 + k_1 k_4) sqrt(xy) the following formula: (A + B) * (C + D) = AC + AD + BC + BD However, even with this remarkably well-boiled down version of your formula, it still -does not reduce- to T=X.Y and your reasoning behind these Kn numbers are quite unclear, in my opinion. Remember, your formula (if your formula theory is to work) will have to work with T=X.Y and the X and the Y have to be something other than 1 and/or T. And the simpler the better. E=MC^2, F=MA, C=PiR^2, etc. Some of the most difficult and fundamental formulas of humankind's history just take a few letters. Think... simple. === Subject: Re: Square roots were the key? something like > T = (3x - 2y)(x+2y) > Seriously, let's take a closer look at this. For the number you want to > factor, T, it is a multiple of two prime numbers, let's call them X and Y. > For the moment, you really don't need all the other information there. So, > upon closer inspection, we can simplify your formula: > T = X . Y > That's the -first- thing you can do to make your life easier. You can even > plug it into your graphing calculator. Let T=15, and let the graphing > function read Y=15/X. When X=3, Y=5. When X=5, Y=3. > See? Easy. Whole numbers are good. BTW, nice -hyperbola-. > so if you solve it out for a factorization you have... > 3=X (or f_1, as you call it) and 5=Y (or f_2) and 15=T. > So after a bit I came up with > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > Above I made your life simpler, now you are making it even more complex, > unnecessarily. Mathematics likes simplicity. No you dim bulb. S here is NOT your target. It's the surrogate. Remarkably now the target is nowhere to be seen. It turns out that if you multiply out, get rid of the sqrt(xy), and then complete the square twice you can find an expression that has only k_1, k_2, k_3, k_4 and S. Now you set the part that has just the k's equal to your target. It's a brilliant idea because now the k's can't be dependent on the factorization of your target, as they're defined by S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) so they're defined relative to the SURROGATE and NOT the target, which clearly escaped you given your reply. The only problem though is that the expression involving the k's is very complicated and has terms with highest exponent of 4. James Harris === Subject: Re: Square roots were the key? > No you dim bulb. What good does name calling do? If you have to resort to name calling, you're not much of an adult IMO. Dave === Subject: Re: Square roots were the key? What good does name calling do? If you have to resort to name calling, > you're not much of an adult IMO. > Dave There's some humor in it, but it's also mean to have SOME shock value. S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) is in fact, I am increasingly sure, the path to surrogate factoring. Being somewhat excited about that I am less controlled in my own postings so I let a few insults fly. But that is against my norm. With that said, I have listened to the complaints and am endeavoring to drop the insults. I do apologize for calling dude a dim bulb, though I do wish he would pay attention to the mathematics and consider why this latest is a brilliant idea! It is a remarkable solution. Such a simple answer right in front of everyone for so long. Just use square roots--and blow the factoring problem apart. James Harris === Subject: Re: Square roots were the key? : No you dim bulb. You're an arrogant asshole for someone who can't get his basic algebra right and has recently (again) admitted that all his self-important pseudo-mathematical ranting got him absolutely nowhere. Justin === Subject: Re: SF: Square roots were the key? : Someone posted a result of that on sci.math where that person used math : software which is a good idea as it's a mess. I was under the impression that you were the most sophisticated and advanced mathematician of the day? If so, why is it that someone else has to use software to work your calculations out for you? : If it can be made to work, it looks like it'll take a lot of expertise : far outside what I usually do, as I focus on rather simple algebra most : of the time. And you screw it up, too. : They kind of just go in a big freaking circle what I call a BFC, where : you need the factorization of T anyway. BFC = Big Freaking Cock-up. Justin === Subject: Re: Square roots were the key? > For years I've been looking for a way to factor one number using the > factorization of some different number, which I call the surrogate for > a method I call surrogate factoring or SF. > Trouble is, every set of equations I played with would either factor > pathetically, as in with a very low success rate, or I'd find that I'd > just go in a circle needing the factorization of my target. And how many times have people been telling you this over and over and over? Two equations and two unknowns doesn't mean squat unless the two equations are independent. What makes you always think that this changes when you have more than two equations? Or just because YOU can't see that they aren't independent? === Subject: minimal anisomorphism I have two directed acyclic graphs which are known to be very nearly isomorphic but are different in a relatively small portion of the graphs. I wish to isolate the anisomorphic bits. Can you suggest any algorithm(s) to do this? I was going to do an edge-by-edge, virtex-by-virtex compare - starting from each point of egress and traversing my way back (in a depth-first-search) to the first miscompare. Then, likewise, starting at each point of ingress and working my way forward to the first miscompare. If each edge and each virtex is 'marked' as it is traversed then, I would submit, that the remaining 'unmarked' set of edges and vertices is the anisomorphism. I believe this will work except for the case when the anisomorphism creates some kind of 'island' of one or more isormorphic subgraphs - but in this particular mapping that is unlikely and would not be critical to my application anyway. === Subject: Two-dimensional correlation analysis information I am to learn two-dimensional correlation analysis and apply it to data sets of time series. I've not had an actual statistics course, but have learned parts through courses in physics. I'd appreciate recommendations on well-regarded texts that explain two-dimensional correlation analysis and provide examples of its use. I'd also appreciate texts on analysis of time series, or time series analysis. === Subject: Why is this true Can someone explain in plain language why the following is true? Any integer is divisible by 3 if the sum of its digits is divisble by 3. === Subject: Re: Why is this true > Can someone explain in plain language why the following is true? > Any integer is divisible by 3 if the sum of its digits is divisble > by > 3. Who is going to post a proof by induction? A little messy, but not toooo bad. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Why is this true >> Can someone explain in plain language why the following is true? >> Any integer is divisible by 3 if the sum of its digits is divisble >> by >> 3. > Who is going to post a proof by induction? A little messy, but not > toooo bad. Induction on the highest power of 10 doesn't look so bad. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Why is this true >>xyzer@hotmail.com > Can someone explain in plain language why the following is true? > Any integer is divisible by 3 if the sum of its digits is divisble by 3. >> Who is going to post a proof by induction? >> A little messy, but not toooo bad. > Induction on the highest power of 10 doesn't look so bad. SIMPLER Casting out nines is a special case x,y = 10,1 of FACTOR THEOREM x-y | P(x)-P(y) in Z[x,y] for poly P in Z[x] This faq is discussed much further in many of my prior posts: --Bill Dubuque === Subject: Re: Why is this true > Can someone explain in plain language why the following is true? > Any integer is divisible by 3 if the sum of its digits is divisble by > 3. Clue: this works in base-10 representation, but not in all others. === Subject: Re: Why is this true > Can someone explain in plain language why the following is true? > Any integer is divisible by 3 if the sum of its digits is divisble by > 3. The powers of 10 all have remainder 1 when divided by 3. === Subject: Re: Why is this true > Can someone explain in plain language why the > following is true? > Any integer is divisible by 3 if the sum of its > digits is divisble by 3. This isn't exactly plain language and is essentially what Stan said in a previous post, but I can't resist. Any integer n may be written as n = d_k 10^k + ... + d_2 10^2 + d_1 10^1 + d_0 10^0 ..where d_i are the digits of n. For instance, the integer 4321 can be expressed as 4321 = 4 * 10^3 + 3 * 10^2 + 2 * 10^1 + 1 * 10. Now if an integer n is divisible by three then we have n == 0 (mod 3) ..or equivalently, n = 0 + 3k for some integer k. Thus, if n is divisble by three then n = d_k 10^k + ... + d_1 10^1 + d_0 10^0 == 0 (mod 3). It follows that 10^m == 1 (mod 3) for any positive integer m since 10 == 1 (mod 3). Hence, if n is divisble by three we have n = d_k 10^k + ... + d_1 10^1 + d_0 10^0 == d_k + ... + d_1 + d_0 == 0 (mod 3), ..which implies the digit sum of n is divisible by three. We can take this whole process the other way and prove that if the digit sum of n is divisble by three, then the integer n is also divisble by three. The same technique can be used to prove a few other similar statements; for example, an integer is divisble by 11 iff the alternating digit sum is divisble by 11. In this example we would have, by assumption, n = d_k 10^k + ... + d_1 10^1 + d_0 10^0 == 0 (mod 11). Now 10^m == (-1)^m (mod 11) for any positive integer m since 10 == -1 (mod 11). Thus, we obtain n = d_k 10^k + ... + d_1 10^1 + d_0 10^0 == (-1)^k d_k + ... -d_1 + d_0 == 0 (mod 11). This establishes the if part of the statement and the converse statement is no harder to prove. A quick example of this would be the integer n = 261206. Since the digit sum 2 - 6 + 1 - 2 + 0 - 6 = -11 == 0 (mod 11) we can conclude that 261206 is divisble by 11. Kyle Czarnecki === Subject: Re: Why is this true > Can someone explain in plain language why the following is true? > Any integer is divisible by 3 if the sum of its digits is divisble by > 3. Hint: Consider 457 = 4*100 + 5*10 + 7 = 4*(99 + 1) + 5*(9 + 1) + 7 = 4*99 + 4 + 5*9 + 5 + 7 = 4*99 + 5*9 + (4 + 5 + 7). Now 99 and 9 are divisible by 3, hence so is 4*99 + 5*9, therefore ... (and the rule of 9's falls out the same way). === Subject: Re: Why is this true >Can someone explain in plain language why the following is true? >Any integer is divisible by 3 if the sum of its digits is divisble by Call numbers congruent modulo 3 iff their difference is divisible by three. This partitions them into three classes: {0,3,6,9,12,...}, {1,4,7,9,13,...}, {2,5,8,10,14,...}. Note that the first set consists of numbers in the form 3k+0, the second in the form 3k+1, and the third in the form 3k+2--in other words, accoriding to their remainded after division by three. Now, let if N = 3n+p and M = 3m+q, then N + M = 3(n+m) + (p+q), i.e., the congruence class (mod 3) of the sum of N and M is the same as that of the sum of their remainders. Furthermore, NM = (3n+p)(3m+q) = 3(3nm+nq+mp) + pq, so that the congruence class of their product is the same as that of the product of their remainders as well. In particular, every power of 10 ({1,10,100...}) has a remainder of 1 when divided by 3, so the class of 10^k * A is the same as that of A. As previously, sums do not change congruence class, so a number expanded in digits (sums of multiples of powers of 10) N = Sum[ N_k 10^k ] = Sum[ N_k ] (mod 3). Hence, the remainder of a number after division by three is the same as the remainder for the sum of its digits. The trick is repeatable for large numbers, for which the sum of digits may itself be large. Since every power of 10 also has remainder 1 when divided by 9, the exact same trick works for division by 9 as well. Additionally, every power of 10 is congruent to either 1 or -1 (=10) modulo 11, so when dividing by 11, an alternating sum of digits with the ones-place digit positive works. For example, if N = 3846, then -3+8-4+6 = 7, so the remainder of 3846/11 is 7. --- Stan Liou === Subject: general proof question For most proofs we do a sequence of logical steps to reach the conclusion. My question is, what is the guarantee that such a sequence of steps will always exist? Is it possible that for our algebra and analysis rules/knowledge, it is impossible to prove something (which experiment indicates is probably true)? === Subject: Re: general proof question >For most proofs we do a sequence of logical steps to reach the conclusion. >My question is, what is the guarantee that such a sequence of steps will >always exist? Is it possible that for our algebra and analysis >rules/knowledge, it is impossible to prove something (which experiment >indicates is probably true)? Certain particular systems, e.g., geometry (at least the Hilbert formulation and some others), have a guarantee that every well- formed statement is decidably true or false, with a proof of each true statement existing. In _general_ however, there is no such guarantee. Not only that, but sometimes there is a gurantee of impossibility. Independence results are actually quite common in mathematics, particularly set theory. Usually, what happens afterward is that one would have two or more mutually contradictory systems. A particularly famous result along these lines is G.9adel's incompleteness theorem, which states that the existence of independent formulas is inevitable for a certain class of mathematical systems. --- Stan Liou === Subject: Re: Help with solving a simple black box of two inputs... boundary=----=_NextPart_000_0012_01C68E2D.93E8AE10 --------------------------------------------------------------------- Just FYI, the function approximates the crude protein content of a pasture based on the average herbage digestibility and proportion of legumes. That last one was for temperate pasture. I'm hoping that you guys could figure out just one more, this time for tropical pasture: x y| 0.0 0.2 0.4 0.6 0.8 1.0 -----+----------------------------------- 0.5 | 7.0 7.5 7.9 8.4 8.9 9.3 0.55 | 9.6 10.2 10.8 11.5 12.1 12.7 0.6 | 12.3 13.1 13.9 14.7 15.5 16.3 0.65 | 15.1 16.1 17.1 18.1 19.1 20.1 0.7 | 18.0 19.2 20.4 21.6 22.8 24.0 The values are quite close to the temperate ones, but of course I need to be as accurate as possible. Note that x in this case has an upper limit at 0.7. Actually, what would be ideal (and probably quite hard to figure out) is a single function of three variables f(x,y,z) that covers both cases. In the previous set of data (temperate), z=0. In the above set (tropical), z=0.16. This is a bit of a long shot, but I was hoping somebody would be able to help me figure out the inner workings of a function of two variables, to which I don't have the implementation. The following table is the input/output map that I've built by profiling it. x y| 0.0 0.2 0.4 0.6 0.8 1.0 -----+----------------------------------- 0.5 | 7.0 7.5 7.9 8.4 8.9 9.3 0.55 | 9.6 10.2 10.9 11.5 12.1 12.8 0.6 | 12.3 13.2 14.0 14.8 15.6 16.4 0.65 | 15.2 16.3 17.3 18.3 19.3 20.3 0.7 | 18.3 19.6 20.8 22.0 23.2 24.4 0.75 | 21.6 23.0 24.5 25.9 27.3 28.8 0.8 | 25.0 26.7 28.3 30.0 31.7 33.3 Unfortunately the outputs are all rounded to a single decimal place. I know the equation must be quite simple, but I don't know what techniques to use to get there. Varying y seems to cause a linear change in the output, however varying x seems to be exponential. Any help appreciated. Nathan === Subject: Re: Help with solving a simple black box of two inputs... <448cef29$0$7192$5a62ac22@per-qv1-newsreader-01.iinet.net.au Just FYI, the function approximates the crude protein content of a pasture based on the average herbage digestibility and proportion of legumes. > That last one was for temperate pasture. I'm hoping that you guys could figure out just one more, this time for tropical pasture: > x y| 0.0 0.2 0.4 0.6 0.8 1.0 > -----+----------------------------------- > 0.5 | 7.0 7.5 7.9 8.4 8.9 9.3 > 0.55 | 9.6 10.2 10.8 11.5 12.1 12.7 > 0.6 | 12.3 13.1 13.9 14.7 15.5 16.3 > 0.65 | 15.1 16.1 17.1 18.1 19.1 20.1 > 0.7 | 18.0 19.2 20.4 21.6 22.8 24.0 > The values are quite close to the temperate ones, but of course I need to be as accurate as possible. Note that x in this case has an upper limit at 0.7. f(x,y) = -11.8 + 25.1 x + 25.0 x^2 - 2.67 y + 4.11 x y+11.8 x^2 y > Actually, what would be ideal (and probably quite hard to figure out) is a single function of three variables f(x,y,z) that covers both cases. In the previous set of data (temperate), z=0. In the above set (tropical), z=0.16. No, that's very easy. If f_1(x,y) works for z = 0 and f_2(x,y) works for z = 0.16, just take f(x,y,z) = f_1(x,y) + (f_2(x,y) - f_1(x,y)) z/0.16. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Help with solving a simple black box of two inputs... This is a bit of a long shot, but I was hoping somebody would be able to >help me figure out the inner workings of a function of two variables, to >which I don't have the implementation. >The following table is the input/output map that I've built by profiling it. >x y| 0.0 0.2 0.4 0.6 0.8 1.0 >-----+----------------------------------- >0.5 | 7.0 7.5 7.9 8.4 8.9 9.3 >0.55 | 9.6 10.2 10.9 11.5 12.1 12.8 >0.6 | 12.3 13.2 14.0 14.8 15.6 16.4 >0.65 | 15.2 16.3 17.3 18.3 19.3 20.3 >0.7 | 18.3 19.6 20.8 22.0 23.2 24.4 >0.75 | 21.6 23.0 24.5 25.9 27.3 28.8 >0.8 | 25.0 26.7 28.3 30.0 31.7 33.3 >Unfortunately the outputs are all rounded to a single decimal place. >I know the equation must be quite simple, but I don't know what >techniques to use to get there. >Varying y seems to cause a linear change in the output, however varying >x seems to be exponential. > The following formula produces all your results, after rounding results > to one decimal place: > f(x,y) = 10.73 x^2 y + 33.13 x^2 + 6.05 x y - 0.104 y^2 > + 16.94 x - 3.279 y - 9.765 How did you get f(x,y)? Interpolating polynomial for two variables? Least squares? (I suspect the latter.) --- Christopher Heckman === Subject: Re: Help with solving a simple black box of two inputs... -=-=-=-=-=- > >This is a bit of a long shot, but I was hoping somebody would be able to >help me figure out the inner workings of a function of two variables, to >which I don't have the implementation. >The following table is the input/output map that I've built by profiling it. > >x y| 0.0 0.2 0.4 0.6 0.8 1.0 >-----+----------------------------------- >0.5 | 7.0 7.5 7.9 8.4 8.9 9.3 >0.55 | 9.6 10.2 10.9 11.5 12.1 12.8 >0.6 | 12.3 13.2 14.0 14.8 15.6 16.4 >0.65 | 15.2 16.3 17.3 18.3 19.3 20.3 >0.7 | 18.3 19.6 20.8 22.0 23.2 24.4 >0.75 | 21.6 23.0 24.5 25.9 27.3 28.8 >0.8 | 25.0 26.7 28.3 30.0 31.7 33.3 > >Unfortunately the outputs are all rounded to a single decimal place. >I know the equation must be quite simple, but I don't know what >techniques to use to get there. >Varying y seems to cause a linear change in the output, however varying >x seems to be exponential. > The following formula produces all your results, after rounding results > to one decimal place: > f(x,y) = 10.73 x^2 y + 33.13 x^2 + 6.05 x y - 0.104 y^2 > + 16.94 x - 3.279 y - 9.765 > How did you get f(x,y)? Interpolating polynomial for two variables? > Least squares? (I suspect the latter.) Linear programming (for best l_infinity approximation). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Difficult limit question from an analysis qual I wasn't able to do this problem and wonder if anyone else has any more insight. The problem is: Find the limit of m * sum(i=m,...,infty) 1/i^2, as m -> infty I tried to figure out an appropriate measure and bounding function that I could use the dominated convergence theorem to allow pulling the limit inside the summation, but no luck. The most obvious sort of measure would be one where integral(f) = f(1)+f(2)/2+...+f(k)/k^2+... in which case our sequence would be {m Xi_{x >= m}} (where Xi means characteristic function). But then I can't find an integrable bound function, and similar remarks go for other kinds of measures. And of course in all this I'm blindly assuming (for sake of heuristics) that the limit actually is 0... could also be infty, or even something in between.. S.P. === Subject: Re: Difficult limit question from an analysis qual > I wasn't able to do this problem and wonder if anyone else has any >more insight. The problem is: >Find the limit of m * sum(i=m,...,infty) 1/i^2, as m -> infty Compare sum(i=m,...,infty) 1/i^2 to sum(i=m,...,infty) 1/(i(i+1). Mike Guy === Subject: Re: Difficult limit question from an analysis qual > I wasn't able to do this problem and wonder if anyone else has any > more insight. The problem is: > Find the limit of m * sum(i=m,...,infty) 1/i^2, as m -> infty Hint: sum(i=m,...,infty) 1/i^2 is between int_[m,oo) dx/x^2 and int_[m-1,oo) dx/x^2 === Subject: Matrix Can you tell me what a matrices is.I do not want a definition but an understanding.If I ask you what are numbers then you wont be abble to tell me accurately but still you feel you know it.Similarly can you tell me what do you mean by matrices? === Subject: Re: Matrix > .... > Can you tell me what a matrices is.I do not want a definition but an > understanding.... That's a good question. Text-books are a bit inclined to drown you in technical details before you can develop a feeling for what matrices are all about. There's an old but very good little book by W. W. Sawyer, Prelude to Mathematics, whose Chapter 8 (pp. 103-124) aims to help you feel at home with matrices in a way which is elementary but also makes good mathematical sense. Ken Pledger. === Subject: Re: Matrix > Can you tell me what a matrices is.I do not want a definition but an > understanding.If I ask you what are numbers then you wont be abble to > tell me accurately but still you feel you know it.Similarly can you > tell me what do you mean by matrices? Matrices provide a convenient way to codify systems of linear equations. So you can write a system as Ax=b where A is an n by n matrix, and x is an unknown column vector, and b is a known column vector. This notation becomes extremely useful when you start doing more complicated things, for example, Gaussian elimination can be rewritten as the so called LU factorization. It gives you a really good sense of equations again and again. Later you can do systems of linear differential equations like dx/dt = Ax and you can write the solution conveniently as x = e^{tA} x(0). Then if you study eigenvalues and eigenvectors and Jordon canonical forms of matrices, this gives you tremendous insight into how the solutions behave. Later you can generalize this to infinite dimensional analogues of these things, and then you can study partial differential equations using the insight gained from studying matrices. === Subject: Re: Matrix In my opnion, matrix is just a set of numbers alined in a rectangle. It can be explained as many kinds of things for the need of science and engineering. The matrix is useful, because of its concision and mathematical properties. Through the matrix, many problems are changed into a translation of matrix spaces. > Can you tell me what a matrices is.I do not want a definition but an > understanding.If I ask you what are numbers then you wont be abble to > tell me accurately but still you feel you know it.Similarly can you > tell me what do you mean by matrices? === Subject: Re: Matrix > Can you tell me what a matrices is.I do not want a definition but an > understanding.If I ask you what are numbers then you wont be abble to > tell me accurately but still you feel you know it.Similarly can you > tell me what do you mean by matrices? Matrix is mother of all living! === Subject: Matrix Can you tell me what a matrices is.I do not want a definition but an understanding.If I ask you what are numbers then you wont be able to tell me accurately but still you feel you know it.Similarly can you tell me what do you mean by matrices? === Subject: Matrix Can you tell me what a matrices is.I do not want a definition but an understanding.If I ask you what are numbers then you wont be able to tell me accurately but still you feel you know it.Similarly can you tell me what do you mean by matrices? === Subject: Re: Matrix- Latin: matrix (singular), matrices (plural): stem (of a tree); metaphorically: the starting point for all kinds of activities or phenomena. In the case of systems of linear equations, all knowledge stems from the table of coefficients and RHS numbers, hence the technical term matrix for rectangular tables of numbers. In geology, matrix is the technical term for the material that encloses isolated pieces of different material. The word matrix is closely related to mater = mother. http://en.wikipedia.org/wiki/Matrix http://en.wikipedia.org/wiki/Matrix_%28mathematics%29 http://en.wikipedia.org/wiki/Determinant It is interesting to observe that the term matrix (coined in 1848 by Sylvester) originated so much later than the term determinant (Leibniz, 1690s). Ciao: Johan E. Mebius >Can you tell me what a matrices is.I do not want a definition but an >understanding.If I ask you what are numbers then you wont be able to >tell me accurately but still you feel you know it.Similarly can you >tell me what do you mean by matrices? === Subject: Re: ELO system > hey, I'd like to apply Prof. Elo's system in a multi-player game, > is that possible? Does anyone knows the right way to do it? You might want to check out the table-tennis system: http://www.ratingscentral.com/ -- Hans Aberg === Subject: Anyons, Ghosts & World Holograms I have been reading Quantum fields in curved space by Birrell & Davies as well as Frank Wilczek's Fractional Statistics and Anyonic Superconductivity. The quantization of gauge forces, EM, weak & strong is not simple (gauge-fixing terms + ghosts) and curved space-time demands the addition of the 3D+1 cousins to 2D+1 anyons called ghosts i.e. fermionic spin 0 scalar fields that violate the spin-statistics connection in a direct physical way not found in globally flat quantum field theory. This is related to the fact you can subtract out zero point energy ZPE in flat QFT, but not in curved QFT - a great error Puthoff makes BTW violating the equivalence principle. It's the strong direct curving of space-time by ZPE that enables the metric engineering of warp and wormhole. Therefore, Hal & Co. (Eric Davis et-al) have literally thrown the baby out with the bathwater. More on this anon. p. 88 Birrell & Davies eqs. 3.192 - 3.195 Note that anyons have fractional quantum statistics as well as fractional angular momentum and are 2D spatial structures. What anyons and ghosts have in common are new forms of quantum statistics. Anyons may be a boundary of ghosts in the sense of the world hologram - this is less than half-baked at the moment. the vacuum polarization cloud ... attaches magnetic flux to charges, and electric charge to flux points in general transmuting the statistics of both ... a vortex carrying the fundamental flux unit 2pi/e acquires fractional electric charge e/2. It also has a total angular momentum 1/4 + integer, and obeys half-fermi statistics. Wilczek p. 48 The traditional spin-statistics connection has no fractional anyonic statistics and demands that spin 0 scalars be bosons not fermions. Violate this and get signal nonlocality punching through event and Ordinary 3D fermions & bosons come from the permutation group. 2D anyons come from the braid group and something called Chern-Simons theory. We're getting there. === Subject: Mathematics to define species using Genomics Re: defining species and rate of speciation Re: Rate of Reassortment <249c82hjvjsnn7copfc9qn8mjo1hajdpuh@4ax.com> <1s6f8214spe29iibcipqfbt8avkaoi53cn@4ax.com> species and will accomodate the concept of species to viruses and >bacteria. That is progress, Bob, and you seem to want to not accept >progress and stick with the past. > I said no such thing. > You spent most of your msg here completely misrepresenting what I > said. You repeated it over and over, saying nothing true or of > substance. Please read messages before replying to them. I won't > bother responding if you falsify what I say next time. You have some > interesting ideas, as well as some weird ones (which is fine). I said > no such thing about use of genomics. Use of genomics in speciation is > old news; more advanced, more subtle tools -- more advanced genomics > -- are available. sentence is not quoting you, but asking for your attention. I sometimes get the feeling of not talking directly to the respondent and so I add his name sometimes in a sentence. But if you read that as a quote, then of course you may get angry and think I am misrepresenting you. For I never misrepresented you or intended on such a thing. I am saying that we need a new concept for species that includes microbes, and you are countering my proposals. So if you continually counter, then of course I can rightfully say you are stuck in the past. For I maybe wrong some of the time, but not wrong all of the time for which your countering posts seem to suggest. Your next three paragraphs sound more like anger than a objective discussion. I care only for objective discussion. > The topic of discussion is microbes (esp viruses). You noted -- > correctly -- that the whole notion of species is much less useful with > microbes. I said to keep Neandertal out this, simply because they are Why do you misspell Neanderthal. I need to have the full Genome of every DNA and RNA that exists on Earth in view of a Genomic definition of Species. Human A,C,G,T is a 3 billion sequence. I understand some plants have the longest sequence of perhaps 20 billion sequence string (am I accurate with that number?) So now a mathematician looking at A,C,G,T sequence of every DNA and RNA on Earth, and finds that a variance in the human sequence of 3 base pairs over a 20,000 sequence is a different species than humans. So is 3/20,000 the minimum variance for all species, comparatively? Now, if virus A comparative to virus B has a variance of 3/20,000, then the mathematician would rightfully say, that Virus A is a different species from Virus B and has different phenotypic behaviour or traits displayed in the environment where these viruses reside. Mathematicians would make a definition of Species more precise than what the biologists have at this moment in time, because (1) A,C,G,T is a number (2) all DNA and RNA have this A,C,G,T (3) it is a sequence (4) sequence varies (5) if it varies enough between two sequences then they are different species So that the old way of defining Species as reproductive viability is only a fraction of the better definition of species as genomic variance. The new way captures reproductive viability but also captures properties or behaviour of DNA/RNA in the environment. A H5N1 that is human to human transmissive has a different environmental behaviour from that of H1N1 and thus a different species of virus. Humans are a different species from apes because they cannot reproduce between them, but also, they are different species because some 8 to 10 million years ago an ape-like ancestor had the behaviour of continually throwing rocks and stones. The variance between Chimpanzees or other apes from the human genome is basically a difference in throwing behaviour. The sum total of the variance in human genome compared to chimpanzee genome is mostly (I would say 90%) because of stonethrowing. That variance is due to the changes of bone and muscle to be able to throw with accuracy and with will and determination and proclivity. The old definition does not capture behaviour. Show me two different species that behave identically the same. There are none, and that is because the variance in the genome sequence. Of course for plants it is something, call it properties, not behaviour. > not microbes. It is of no relevance here that genomics has contributed > to the discussion of Neandertal (though rather slightly; only very > limited genomic data is available), and of course to many other higher > organisms, as you note later. > Genomics will contribute, in a sense even more, to microbial work. But > that doesn't get at the key underlying problem, with the nature of > microbes and microbial variability. There is no particular reason that > the notion of species must be applied there at all. Certainly attempts A new tool opens up things that we cannot forsee now. > to do so so far have been very mixed, and the genomics info has made > the situation worse in some ways (by highlighting the variability). > The terms are not what is important; the nature of the organisms is. > The term is useful only if it conveys a meaning. This is more than terms and definitons and concepts. This is about understanding of how biology works. And likewise, a term is harmful if it conveys a fuzzy meaning or is contradictory or counter to the biological data. So our old definition of species is now harmful to understanding biology in light of progress of genome sequencing. > With flu virus, genomic info should be of great interest; I just > referred you to a paper on that. The problem is not with the use of > genomic info but with an attempt to talk about species. Rather than > discussing the abstraction of species, we should be discussing the > properties of flu virus. Unfortunately, the key characteristics of the > various kinds of flu virus are rather unclear, though some differences > and some likely virulence factors are noted. There is no reason to > question that they all are the same species (all influenza A, at > least); more importantly, what we call them is not the concern. I agree with you that a definition is just a desire to make things clear. But I disagree as to the usefulness of a definition. For you mention property which I deem as behaviour. And it is the variance of the genomic sequence that gives rise to this property. So to say that H1N1 compared to H5N1 has a genomic variance of 3/20,000 means that they are two different species of virus with 2 different properties or environmental behaviours. So that if the recent family deaths in Indonesia by H5N1 were tested by genome project and found to vary from H1N1 not by 3/20,000 but by 4/20,000 is an indicator that this virus is soon to be a new species from H5N1 and with possible human to human transmission. You see the definition of species becomes this number-- the rate of variance over the sequence, the 3/20,000 (if that is accurate) becomes the definition of species. Here is an example of where the definition of species becomes pivotal. If this number 3/20,000 is accurate over a large class of plants and animals then raises interesting theoretical questions of given a animal A such as humans and scrambling their genome just shy of the 3/20,000 mark. All of those DNA sequences that are less than the 3/20,000 mark, would every one of them still be humans? You see, a genomic definition opens up brand new doors to future progress. > By the way, the issue of Science I aimed you to also has in it one of > track; I understand that their basic findings have been confirmed by > others. Some would argue that growth of the current bird flu virus > deep in the respiratory track helps explain why it is not easily > transmitted human-human (ie, it doesn't get out very well). It may > also provide a reason why the bird flu won't recombine/reassort with > the human virus, at least for now -- if they are growing in different > cells. > bob upper part of the respiratory system have a 3/20,000 variance from viruses that inhabite the lower track which are different lung cells. Species becomes this rate number, and species includes behaviour of the virus, or as Bob calls it, property. (Not a misrepresentation when I add your name, Bob, into a sentence), so cool down and just objectively discuss, where anger is out of place here. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Mathematics to define species using Genomics Re: defining species and rate of speciation Re: Rate of Reassortment <249c82hjvjsnn7copfc9qn8mjo1hajdpuh@4ax.com> <1s6f8214spe29iibcipqfbt8avkaoi53cn@4ax.com> Ross-c === Subject: Science world cup Is Newton a better dribbler than Einstein? Is Celcius too hot for Curie? Would Faraday's electric pace be too much for Becquerel? Find out on the Science World Cup wallchart, free at www.null-hypothesis.co.uk Who says science and football don't mix... === Subject: Ulam's Rose nailed down every possible prime to fit in SEVEN linear equation. By using different configurations of Ulams rose. And that his superior understanding of primes actually originated from studying these 7 equations? But the minims saw to that the information about the functions was supressed because the formulas was thought to have divine information from the creator? Some people still have information about these equation, and play games with us writing screenplayes, like in the name of the rose. And the who tried to release these supressed equations. Allegidbly Ulams rose with it's origin within the golden ratio and studies in the proportion of the living species. It is all about Phi ;) Any truth to this story or yet another internet legend (or did i just started one lol ;D ? === Subject: Ulam's Rose nailed down every possible prime to fit in SEVEN linear equation. By using different configurations of Ulams rose. And that his superior understanding of primes actually originated from studying these 7 equations? But the minims saw to that the information about the functions was supressed because the formulas was thought to have divine information from the creator? Some people still have information about these equation, and play games with us writing screenplayes, like in the name of the rose. And the who tried to release these supressed equations. Allegidbly Ulams rose with it's origin within the golden ratio and studies in the proportion of the living species. It is all about Phi ;) Any truth to this story or yet another internet legend (or did i just started one lol ;D ? === Subject: Re: Ulam's Rose Would there be any implifications for factorisation RSA prime products if these equations for establishing a number is prime in just seven tests existed? jt64@tele2.se skrev: > nailed down every possible prime to fit in SEVEN linear equation. By > using different configurations of Ulams rose. > And that his superior understanding of primes actually originated from > studying these 7 equations? > But the minims saw to that the information about the functions was > supressed because the formulas was thought to have divine information > from the creator? > Some people still have information about these equation, and play games > with us writing screenplayes, like in the name of the rose. And the > who tried to release these supressed equations. Allegidbly Ulams rose > with it's origin within the golden ratio and studies in the proportion > of the living species. It is all about Phi ;) > Any truth to this story or yet another internet legend (or did i just > started one lol ;D ? === Subject: Re: Ulam's Rose I can see it would be easy to precompute large primes like used in the with the equations RSA but otherwise? Would it even be possible to store them? How many primes are their to expect in the last current noneressolved RSA products of today? Would they be storable and how much faster could one factor if one had possibility to store them all. Noone knows exact of course but in the range of their estimated size? j...@tele2.se skrev: > Would there be any implifications for factorisation RSA prime products > if these equations for establishing a number is prime in just seven > tests existed? > jt64@tele2.se skrev: > nailed down every possible prime to fit in SEVEN linear equation. By > using different configurations of Ulams rose. > And that his superior understanding of primes actually originated from > studying these 7 equations? > But the minims saw to that the information about the functions was > supressed because the formulas was thought to have divine information > from the creator? > Some people still have information about these equation, and play games > with us writing screenplayes, like in the name of the rose. And the > who tried to release these supressed equations. Allegidbly Ulams rose > with it's origin within the golden ratio and studies in the proportion > of the living species. It is all about Phi ;) > Any truth to this story or yet another internet legend (or did i just > started one lol ;D ? === Subject: Re: Ulam's Rose > I can see it would be easy to precompute large primes like used in the > with the equations RSA but otherwise? > Would it even be possible to store them? How many primes are their to > expect in the last current noneressolved RSA products of today? > Would they be storable and how much faster could one factor if one had > possibility to store them all. > Noone knows exact of course but in the range of their estimated size? > j...@tele2.se skrev: There are approximately 10^80 atoms in the entire universe, out to the radius where the recession velocity reaches the speed of light. The number of primes that could be factors of currently-practised RSA moduli is enormously greater. -- === Subject: Re: Ulam's Rose Would not rsa 576 have 212 decimal digits and if we assume symetrical size of the factors they would have size of 10^106. I have no idea how many of these actually could be prime, but i HIGHLY doubt it could be 10^80. How many percent of numbers is prime if max 10^120 would not that give a more real world connected picture? I have no idea about the prime distribution, i would love to see a graph how they decrease anyone got one? bert skrev: > I can see it would be easy to precompute large primes like used in the > with the equations RSA but otherwise? > Would it even be possible to store them? How many primes are their to > expect in the last current noneressolved RSA products of today? > Would they be storable and how much faster could one factor if one had > possibility to store them all. > Noone knows exact of course but in the range of their estimated size? > j...@tele2.se skrev: > There are approximately 10^80 atoms in the entire universe, > out to the radius where the recession velocity reaches the > speed of light. The number of primes that could be factors > of currently-practised RSA moduli is enormously greater. > -- === Subject: Re: Ulam's Rose jt64@tele2.se a .8ecrit : > Would not rsa 576 have 212 decimal digits and if we assume symetrical > size of the factors they would have size of 10^106. > I have no idea how many of these actually could be prime, but i HIGHLY > doubt it could be 10^80. You are right, it is MUCH higher (around 4.10^103) > How many percent of numbers is prime if max 10^120 would not that give > a more real world connected picture? > I have no idea about the prime distribution, i would love to see a > graph how they decrease anyone got one? A thing called Prime number theorem could help you ; the percentage you are looking for is roughly 1/ ln(n) > bert skrev: > I can see it would be easy to precompute large primes like used in the > with the equations RSA but otherwise? > Would it even be possible to store them? How many primes are their to > expect in the last current noneressolved RSA products of today? > Would they be storable and how much faster could one factor if one had > possibility to store them all. > Noone knows exact of course but in the range of their estimated size? > j...@tele2.se skrev: >> There are approximately 10^80 atoms in the entire universe, >> out to the radius where the recession velocity reaches the >> speed of light. The number of primes that could be factors >> of currently-practised RSA moduli is enormously greater. >> -- === Subject: Re: Ulam's Rose > Would not rsa 576 have 212 decimal digits and if we assume symetrical > size of the factors they would have size of 10^106. > I have no idea how many of these actually could be prime, but i HIGHLY > doubt it could be 10^80. > How many percent of numbers is prime if max 10^120 would not that give > a more real world connected picture? > I have no idea about the prime distribution, i would love to see a > graph how they decrease anyone got one? > bert skrev: > There are approximately 10^80 atoms in the entire universe, > out to the radius where the recession velocity reaches the > speed of light. The number of primes that could be factors > of currently-practised RSA moduli is enormously greater. > -- No idea and highly doubt do not give us an answer, but the prime number theorem does: there are approximately 3.6 * 10^117 primes which are less than 10^120. Some of them would be unsuitable for RSA, e.g. those with less than 80 digits, or where P+1 or P-1 is highly composite, but they comprise an insignificant fraction of the total. -- === Subject: New mathematics/physical sciences positions at http://jobs.phds.org, June 12, 2006 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Scientific Programmer/Analyst: Quantitative Financial Strategies, Inc., Greenwich, CT. This entry-level position (0-2 years experience) in our Research Department represents an excellent opportunity for an individual to gain both knowledge of financial market data and exposure to sophisticated statistical analysis. The... http://jobs.phds.org/jobs/qfsfunds/scientific * Assistant Portfolio Manager/Portfolio Manager (6/7/06): Aperio Group LLC, San Francisco, CA. Assist senior portfolio manager in management of structured, tax-managed equity portfolios and assist Chief Investment Officer on special projects. Success in this role will lead to the position of Portfolio Manager. The hours for this http://jobs.phds.org/jobs/psolli/assistant-portfolio * Excel VBA Developer IRD: Phi Partners, London. This Bank's Rates group offers a challenging position as a VBA Developer that requires both business and technical skills to build a spreadsheet-based trade capture application. Three plus years of VBA development experience along with strong experience in fixed income or... http://jobs.phds.org/jobs/phipartners/excel-vba-developer * Time Series Analysis: Paul & Dominic, London or NY. We have a bank looking for PhDs with a deep understanding of signal processing, filters, forecasting, time series analysis etc. The work is to look at price data and determine the signals to be found within it. This is extremely noisy, non bandwidth limited, and... http://jobs.phds.org/jobs/DominiConnor/time-series-analysis * San Francisco Bay Area Investment Firm Looking for Developer with Quantitative Background: Redwood, San Francisco Bay Area, CA. The primary focus of the position is on improving and maintaining existing trading related applications. The ability to conduct research and perform analysis will be required for ad-hoc projects. Specific project assignments will depend on candidate's skills and... http://jobs.phds.org/jobs/redwoodsf/san-francisco-bay * Derivatives Quantitative Analyst: Capital Markets Recruiting Partners, Inc., Chicago, IL. 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This is an exciting... http://jobs.phds.org/jobs/dinkak/bond-pricing-analyst * PhD position in Soil-water plant interactions: Forschungszentrum Juelich GmbH, Juelich, Germany. For our Institute of Chemistry and Dynamics of the Geosphere - Agrosphere Department (ICG-IV) - we are seeking a PhD Fellow (Soil-water plant interactions) for a three-year PhD project Tasks: The candidate will develop his/her research in the field... http://jobs.phds.org/jobs/javaux/phd-position-in-soil * Risk Management Quant in a Universal Bank: BCV, Lausanne,CH. I am looking for a candidate to join our market risk management team. The ideal candidate would have experience in model implementation, solid skills in mathematical modelling for derivatives pricing, and excellent programming capabilities. EU citizens. The position requires a... http://jobs.phds.org/jobs/susinno/risk-management * Post Doctoral Fellow in Mathematical Biology: INRA, France. A two-years post-doctoral fellow position is available at the Laboratory of Applied Mathematics and Computer Science ( http://www.jouy.inra.fr/ unites/miaj/rubrique.php3?id_rubrique=184 ) at the French National Institute for Agricultural Research ( http://www.international.inra.fr/ ).... http://jobs.phds.org/jobs/JWang/post-doctoral-fellow * Quantitative Strategist : Hagan-Ricci Group, NYC. NYC electronic brokerage is seeking to add a Junior Quantitative Strategist to its algorithmic trading team. Work closely with developers, traders, and the product manager for the algorithms. This is an opportunity to be part of the development of equity trading algorithms... http://jobs.phds.org/jobs/HRG/quantitative * Exceptional Software Engineer -boutique finance firm: The Dauren Harris Group (search firm), NYC. Extraordinary computer scientists wanted. You have an impeccable academic track record with at least one degree in computer science, and a strong desire to work in the fast-paced financial industry. This position demands a true focus and love of computer... http://jobs.phds.org/jobs/lrosenberg/exceptional-software === Subject: Re: Convergence problem Yes, but why is it that my series expansion doen't work? I'm still at lost what the flaw is in the original paper. But working along another line of thought has led to a satisfactory solution now, which will be published soon on the web. Han de Bruijn === Subject: Murder in the name of the rose Would any organisation other than religious murder to supress mathematical information? If so who and why? MrPerfect === Subject: Re: Murder in the name of the rose >Would any organisation other than religious murder to supress >mathematical information? The closest thing I know of to a case of murder to suppress mathematical information, whether religious or not, is the story of Hippasus being killed by the Pythagoreans. But that version of the story is modern, possibly originating with Morris Kline (see ). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Murder in the name of the rose > Would any organisation other than religious murder to supress > mathematical information? > If so who and why? I wouldn't exclude the possibility of a government murdering someone who might be able to find a way to break their cryptographic codes. Jose Carlos Santos === Subject: Re: Murder in the name of the rose days. My association with the Department is that of an alumnus. >> Would any organisation other than religious murder to supress >> mathematical information? >> If so who and why? >I wouldn't exclude the possibility of a government murdering someone who >might be able to find a way to break their cryptographic codes. If the way is the good old rubber hose cryptoanalysis[1], then perhaps. But if the way means that someone found a way to break them analytically (think Rejewski, Turing, et al with ENIGMA, or Friedman et al with PURPLE), then I find it hard to believe. Mostly, because such a murder would not make the system secure. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu [1] I.e., beating up someone who has the key in order to obtain the key. === Subject: Murder in the name of the rose Would any organisation other than religious murder to supress mathematical information? If so who and why? MrPerfect === Subject: Superposition of solutions in PDF Could you please help me out with the following? I am solving: aP''+bP**+cP'=0 where ' and * denote partial derivatives of P(x,y) with respect to the independent variables and a,b,c are constants. I first find a separable solution for P=X(x)Y(y), which are a sum of exponentials [X(x)]and a sum of cos/sin [Y(y)] respectively. Then, how does one superpose for a general solution? Does one integrate X,Y over the separation constant, i.e. int_{0}^{infty} dk X(x) Y(y)? Lenore === Subject: Re: Superposition of solutions in PDF >Could you please help me out with the following? >I am solving: aP''+bP**+cP'=0 where ' and * denote partial >derivatives of P(x,y) with respect to the independent variables and >a,b,c are constants. I first find a separable solution for P=X(x)Y(y), >which are a sum of exponentials [X(x)]and a sum of cos/sin [Y(y)] >respectively. Then, how does one superpose for a general solution? Does >one integrate X,Y over the separation constant, i.e. int_{0}^{infty} dk >X(x) Y(y)? Hint: are you given boundary conditions? If so, use them. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: A basic question on real numbers I have the following basic question:- Irrational numbers can be obtained as the limit of cauchy sequence of rational numbers. Is there a proof that a convergent infinite series having irrational terms converges to a number which can always be obtained as the (above) limit of cauchy sequence of rationls. I would like to see the proof for this statement. === Subject: Re: A basic question on real numbers > I have the following basic question:- > Irrational numbers can be obtained as the limit of > cauchy sequence of rational numbers. > Is there a proof that a convergent infinite series > having irrational terms converges to a number which > can always be obtained as the (above) limit of cauchy > sequence of rationls. > I would like to see the proof for this statement. Let {x_n} converge to x, where each x_n is irrational. For each n, choose a sequence {r_n_m} of rationals that converges to x_n. Then the diagonalization of these sequences of rationals -- namely, the sequence r_1_1, r_2_2, r_3_3, ... -- is a sequence of rationals that converges to x. This isn't immediately evident (to a beginner, at least), but the proof also isn't very difficult. More generally (same proof method), a limit of limits is a limit, or in other language, the (sequential) closure of a set B can be obtained by (one application of) collecting together all the limits of convergent sequences in B. Dave L. Renfro === Subject: Re: A basic question on real numbers >I have the following basic question:- >Irrational numbers can be obtained as the limit of cauchy sequence of >rational numbers. >Is there a proof that a convergent infinite series having irrational >terms converges to a number which can always be obtained as the (above) >limit of cauchy sequence of rationls. >I would like to see the proof for this statement. Well, in fact _any_ real number is the limit of a cauchy sequence of rationals - this has nothing to do with the fact that the number is a coonvergent infinite series having irrational terms. How you prove this depends on where you're starting. I don't know what the context is: (i) if this is a situation where you have some well-defined set of axioms that the real numebers satisfy then tell us what your axioms are and someone will show you the proof from those axioms. (ii) if on the other hand this is all very informal, we haven't said yet exactly what the real numebers are, then consider the following: Probably you believe this: (*) If x is any real number then there exists an integer n such that n <= x < n + 1. We call this integer the floor of x, written [x]. If you believe (*) then it's easy to prove the following theorem: Thm. If x is a real number then there exists a sequence of rationals that converges to x. Proof: First note that for any x we have [x] <= x < [x] + 1, which implies that 0 <= x - [x] < 1, which implies that (*) |x - [x]| < 1. Now suppose that x is a real number, and for n = 1, 2, ... define r_n = [n*x]/n. Then r_n is certainly rational, since [n*x] is an integer. And |x - r_n| = |(n*x - [n*x]) / n| = |n*x - [n*x]| / n < 1/n, by (*). This shows that r_n -> x as n -> infinity. QED. ************************ David C. Ullrich === Subject: Re: A basic question on real numbers |I have the following basic question:- | | Irrational numbers can be obtained as the limit of cauchy sequence of | rational numbers. | | Is there a proof that a convergent infinite series having irrational | terms converges to a number which can always be obtained as the (above) | limit of cauchy sequence of rationls. | | I would like to see the proof for this statement. | Take the limit of the convergent series of irrational terms and express it as a decimal. WOLOG I'm assuming that the number lies in (0,1) -- if not, just add a whole number at the end. Represent this number as X = .d_1d_2... then the absolutely convergent series of rationals Sn = Sum(i,1,n) d_i*10^i converges to the limit X. Norm === Subject: Re: A basic question on real numbers aone1504@yahoo.com nous a r.8ecemment amicalement signifi.8e : > I have the following basic question:- > Irrational numbers can be obtained as the limit of cauchy sequence of > rational numbers. > Is there a proof that a convergent infinite series having irrational > terms converges to a number which can always be obtained as the > (above) limit of cauchy sequence of rationls. > I would like to see the proof for this statement. If u_n is an infinite serie with irrational terms which converges to a number r, then : v_n = [10^n u_n] 10^-n is a infinite serie having only rational terms which converges to the same number r -- Patrick === Subject: Re: A basic question on real numbers > aone1504@yahoo.com nous a r.8ecemment amicalement signifi.8e : > I have the following basic question:- > Irrational numbers can be obtained as the limit of cauchy sequence of > rational numbers. > Is there a proof that a convergent infinite series having irrational > terms converges to a number which can always be obtained as the > (above) limit of cauchy sequence of rationls. > I would like to see the proof for this statement. > If u_n is an infinite serie with irrational terms which converges to a > number r, then : > v_n = [10^n u_n] 10^-n is a infinite series having only rational terms > which converges to the same number r Where [x] is the largest integer not greater than x. Nice! === Subject: Re: A basic question on real numbers > I have the following basic question:- > Irrational numbers can be obtained as the limit of cauchy sequence of > rational numbers. > Is there a proof that a convergent infinite series having irrational > terms converges to a number which can always be obtained as the (above) > limit of cauchy sequence of rationls. > I would like to see the proof for this statement. More generally, the limit of *any* convergent series can always be obtained as the limit of a sequence of rational numbers (there's no need to write Cauchy sequence here, since every convergent sequence is a Cauchy sequence). Let _l_ be the limit. Then _l_ is a real number and *any* real number is the limit of some sequence of rational numbers. Jose Carlos Santos === Subject: differential equation+even Coudl you share you ideas with teh following seemingly interesting problem: Let f be from C^1(R) and u(x) is a solution of the differential equation: u''(x) + f(u(x)) = 0 such that u(x) > 0 for all x in (-1,1) and u(1) = u(-1) = 0. Prove that u(x) is even fucntion. I'm a newbie in differential equation, are there any methods to solve such equation as u''(x) + f(u(x)) = 0 ? Thansk === Subject: Re: differential equation+even >Coudl you share you ideas with teh following seemingly interesting problem: >Let f be from C^1(R) and u(x) is a solution of the differential equation: >u''(x) + f(u(x)) = 0 such that u(x) > 0 for all x in (-1,1) >and u(1) = u(-1) = 0. Prove that u(x) is even fucntion. >I'm a newbie in differential equation, are there any methods to solve >such equation as u''(x) + f(u(x)) = 0 ? Not in general. Hint: let p in (-1,1) such that u'(p) = 0. Show that u(2p-x) = u(x). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: differential equation+even > Coudl you share you ideas with teh following seemingly interesting problem: > Let f be from C^1(R) and u(x) is a solution of the differential equation: > u''(x) + f(u(x)) = 0 such that u(x) > 0 for all x in (-1,1) > and u(1) = u(-1) = 0. Prove that u(x) is even fucntion. > I'm a newbie in differential equation, are there any methods to solve such equation as u''(x) + f(u(x)) = 0 ? > Thansk Let's assume that u(x) is odd. Then u(-x) = -u(x). The second derivative of an odd function is still odd. So if u(x) were odd, then the differential equation would be - u(x) + f (-u(x)) = 0 on the negative side. Since u(x) > 0 everywhere on it's domain, if f were an odd function, this would be equivalent to the original equation. However, in general, f may not be odd, so this would result in a different equation. Hence, the u(x) must be even to preserve the form of the equation on the negative side. === Subject: Re: differential equation+even > Coudl you share you ideas with teh following > seemingly interesting problem: > Let f be from C^1(R) and u(x) is a solution of the > differential equation: > u''(x) + f(u(x)) = 0 such that u(x) > 0 for all x > in (-1,1) > and u(1) = u(-1) = 0. Prove that u(x) is even > fucntion. > I'm a newbie in differential equation, are there > any methods to solve such equation as u''(x) + > f(u(x)) = 0 ? > Thansk > Let's assume that u(x) is odd. I think that you assumption is quite restrictive, we can still have many twice differentiable functions u(x) with this properties which are neither odd, nor even. >Then u(-x) = -u(x). > The second > derivative of an odd function is still odd. So if > u(x) were odd, then > the differential equation would be - u(x) + f > (-u(x)) = 0 on the > negative side. Since u(x) > 0 everywhere on it's > domain, if f were an > odd function, this would be equivalent to the > original equation. > However, in general, f may not be odd, so this would > result in a > different equation. Yes, really f may be far from odd... >Hence, the u(x) must be even to > preserve the form > of the equation on the negative side. Well, under you hypotheses on u and f it is ok, but i think it describes very little class of the possible cases - in my humble opinion, it is just a partcilar case. === Subject: Re: differential equation+even >>Coudl you share you ideas with teh following seemingly interesting problem: >>Let f be from C^1(R) and u(x) is a solution of the differential equation: >>u''(x) + f(u(x)) = 0 such that u(x) > 0 for all x in (-1,1) >>and u(1) = u(-1) = 0. Prove that u(x) is even fucntion. >>I'm a newbie in differential equation, are there any methods to solve such equation as u''(x) + f(u(x)) = 0 ? >>Thansk > Let's assume that u(x) is odd. Then u(-x) = -u(x). The second > derivative of an odd function is still odd. So if u(x) were odd, then > the differential equation would be - u(x) + f (-u(x)) = 0 on the > negative side. Since u(x) > 0 everywhere on it's domain, if f were an > odd function, this would be equivalent to the original equation. > However, in general, f may not be odd, so this would result in a > different equation. Hence, the u(x) must be even to preserve the form > of the equation on the negative side. However, the problem does not appear to preclude the possibility that f be an odd function. Dale === Subject: Cauchy method of residues for double integrals ? I'd like to know if the Cauchy method of residues is still valid for double (or multiple) integrals ? Where could I find info on this subject ? I couldn't find anything about JH === Subject: Re: Cauchy method of residues for double integrals ? > I'd like to know if the Cauchy method of residues is still valid for > double (or multiple) integrals ? Your question is not very precise. Assuming you mean the following: 1. you were able to transform a multiple integral into a product of integrals, all of them disjunct. 2. Some of these integrals can be complexified i.e. replacing a real variable by a complex one. 3. If you a then able to solve these complex integrals be the residue method, the results must be real each. 4. What is the problem then? (B) When calculating multidimensional integrals you should alway try to reduce the dimension by applying Greens theorem and/or stokes theorem and then carry on with the method above. -- Stephan W. M.9fller === Subject: Prime number distribution Could anyone show me how percentage of primenumbers for 10^2 to 10^20, changes for added digit. How big part of the numbers in RSA 576, could actually be primes if we assume no factor bigger than 10^120? Someone told me at least 10^80 i do not think that compute or does it? === Subject: Re: Prime number distribution > Could anyone show me how percentage of primenumbers for 10^2 to > 10^20, changes for added digit. > How big part of the numbers in RSA 576, could actually be primes if we > assume no factor bigger than 10^120? > Someone told me at least 10^80 i do not think that compute or does it? The number of (positive) primes less than or equal to (natural number) n is the prime counting function, often denoted pi(n). Since (according to the Prime Number Theorem) the value of pi(n) is (very approximately) n/ln(n), the density of primes is roughly proportional to 1/ln(n) with increasing n. In other words, prime density is inversely proportional to the number of digits. More along these lines here: http://en.wikipedia.org/wiki/Prime_number_theorem === Subject: Re: Prime number distribution > Could anyone show me how percentage of primenumbers for 10^2 to > 10^20, changes for added digit. > How big part of the numbers in RSA 576, could actually be primes if we > assume no factor bigger than 10^120? > Someone told me at least 10^80 i do not think that compute or does it? Why start a new thread, or seek to clutter the newsgroup with perfectly well-known data? If you visit mathworld.wolfram.com, you will find everything you could possibly absorb about prime numbers and their distribution, both actual and theoretical. -- === Subject: Re: Questions about probability functions Hello Robert, As my time has been here and there on a variety of projects I haven't had much time to put the probability and statistics work to use as much as I would like, but I did want to know the answer to another question or two if you wouldn't mind: 1. I understand that there are several different types of distributions. Two distributions which seem to be more relevant to me than others right now are the normal and the standard distributions. Are these the same thing by different names? 2. Why wouldn't a random distribution simply be a flat line of some value Y over a defined interval [a,b]? In other words, why would it be a bell curve when there is equal likelihood that a random value will be selected over the range [a,b] -- wouldn't any draw from the interval be equally likely to occur and thus there would be no slope? (This is just something I was pondering recently.) 3. What exactly does sigma (standard deviation) really measure, in graphical or plain-English terms? For example, I seem to recall it's the value or point at which something happens on a distribution, but what that something is I can't recall. Seems to me like sigma (1-sigma to be exact) is the point at which the curve values are 0.68 of the max or something like that, but I'm reaching here. Mike >1. In a standard distribution, what more is needed to characterize the >bell curve other than the mean and the standard deviation (sigma)? I >ask this because some of the modeling and analysis tools that I am >using seem only to need those two parameters to determine fragment >dispersions and flyouts. > The mean and standard deviation are the only two parameters for the > normal distribution. Of course, there are other distributions... >2. In plain English, what is the difference between a probability >density function (PDF) and a cumulative distribution function (CDF)? I >know one is the derivative of the other and so on, but what is the >physical significance of the two? > The CDF of a random variable X tells the probability that the random > variable is at most a given value, i.e. F(x) = Prob {X <= x}. > The PDF f(x) says that if you consider a small interval of values > near x, the probability that X is in that interval is approximately > f(x) times the length of the interval. Thus > f(x) is the limit of Prob{x <= X <= x + t}/t or > Prob{x - t <= X <= t}/t as t -> 0+. >3. In Monte Carlo simulation, isn't that basically just a random-number >generator being used for draws against a particular phenomenon, and >then comparing the draw to a scale or thermometer to determine the >outcome of the event in question? > I'm not sure what the question is. In the context of a random variable > arising in some process, the basic idea is to use a pseudo-random number > generator to simulate the process, generating a large sample from (if > you did it right) the distribution of the random variable. You > then study the sample to draw whatever conclusions you want about the > distribution of the random variable. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Questions about probability functions distributions. Two distributions which seem to be more relevant to me > than others right now are the normal and the standard distributions. > Are these the same thing by different names? There are lots of types of distributions. I don't know of any called a standard distribution, but there's a standard normal distribution, which is just the normal distribution with mean 0 and standard deviation 1. > 2. Why wouldn't a random distribution simply be a flat line of some > value Y over a defined interval [a,b]? In other words, why would it be > a bell curve when there is equal likelihood that a random value will be > selected over the range [a,b] -- wouldn't any draw from the interval be > equally likely to occur and thus there would be no slope? (This is > just something I was pondering recently.) If the distribution is uniform on the interval [a,b], in the sense that the probabilities of different sub-intervals of the same length are always equal, then indeed it is not a bell curve: the density of the uniform distribution on the interval [a,b] is constant on that interval. The bell curve is for a normal distribution. > 3. What exactly does sigma (standard deviation) really measure, in > graphical or plain-English terms? For example, I seem to recall it's > the value or point at which something happens on a distribution, but > what that something is I can't recall. Seems to me like sigma (1-sigma > to be exact) is the point at which the curve values are 0.68 of the max > or something like that, but I'm reaching here. The standard deviation is a measure (not the only one, but for various reasons the most commonly used one) of how spread-out the distribution is. If the mean of the random variable X is mu, the standard deviation of X is the square root of the mean of (X - mu)^2. For a normal distribution, the probability of |X - mu| < sigma is approximately 0.6827, but it's different for other distributions. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Questions about probability functions distributions. Two distributions which seem to be more relevant to me > than others right now are the normal and the standard distributions. > Are these the same thing by different names? > There are lots of types of distributions. I don't know of any called a > standard > distribution, but there's a standard normal distribution, which is > just the > normal distribution with mean 0 and standard deviation 1. That is the one I meant. > 2. Why wouldn't a random distribution simply be a flat line of some > value Y over a defined interval [a,b]? In other words, why would it be > a bell curve when there is equal likelihood that a random value will be > selected over the range [a,b] -- wouldn't any draw from the interval be > equally likely to occur and thus there would be no slope? (This is > just something I was pondering recently.) > If the distribution is uniform on the interval [a,b], in the sense that > the > probabilities of different sub-intervals of the same length are always > equal, > then indeed it is not a bell curve: the density of the uniform > distribution on the > interval [a,b] is constant on that interval. The bell curve is for a > normal > distribution. So what would you call a non-bell curve distribution? The picture I have in my head is, one decides that a probability problem can be represented in terms of a flat line of constant value Y across the interval [a,b] and zero outside of it. In other words, all outcomes are equally likely (discrete or continuous, I suppose, it doesn't matter to me). > 3. What exactly does sigma (standard deviation) really measure, in > graphical or plain-English terms? For example, I seem to recall it's > the value or point at which something happens on a distribution, but > what that something is I can't recall. Seems to me like sigma (1-sigma > to be exact) is the point at which the curve values are 0.68 of the max > or something like that, but I'm reaching here. > The standard deviation is a measure (not the only one, but for various > reasons > the most commonly used one) of how spread-out the distribution is. > If the mean > of the random variable X is mu, the standard deviation of X is the > square root of > the mean of (X - mu)^2. For a normal distribution, the probability of > |X - mu| < sigma > is approximately 0.6827, but it's different for other distributions. That seems to make a lot of sense to me now...in physical terms, you are saying that the bell is more tight or spread out depending on the value of sigma. Part of the reason this is of interest to me is that I am working on engagement simulations for fragmenting munitions, each with its own probabilistic dispersions. For example, in polar/arena zone testing, there are (mu,sigma) distributions on things like fragment mass, fragment velocity, etc. and I need a good understanding of what the sigma is for a physical property such as fragment mass (i.e. if the mean mass of a fragment is 1.0 g and the standard deviation is 0.5 g, what does that really mean? etc.) Mike === Subject: Re: Questions about probability functions >> 2. Why wouldn't a random distribution simply be a flat line of some >> value Y over a defined interval [a,b]? In other words, why would it be >> a bell curve when there is equal likelihood that a random value will be >> selected over the range [a,b] -- wouldn't any draw from the interval be >> equally likely to occur and thus there would be no slope? (This is >> just something I was pondering recently.) >> If the distribution is uniform on the interval [a,b], in the sense that >> the >> probabilities of different sub-intervals of the same length are always >> equal, >> then indeed it is not a bell curve: the density of the uniform >> distribution on the >> interval [a,b] is constant on that interval. The bell curve is for a >> normal >> distribution. >So what would you call a non-bell curve distribution? The picture I >have in >my head is, one decides that a probability problem can be represented >terms of a flat line of constant value Y across the interval [a,b] and >zero outside >of it. In this case you'd call it a uniform distribution on the interval [a,b]. > In other words, all outcomes are equally likely (discrete or >continuous, I suppose, >it doesn't matter to me). But it does matter to me. This is a continuous distribution, and for any continuous distribution all individual outcomes have probability 0. So all outcomes are equally likely is not useful here. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Another question for you - sample spaces Hi Robert, Something else I've been wondering about lately and can't recall the answer to relates to polling and sample sizes. I seem to recall some theorem stating that if you want to poll a sample size of S that it is not necessary to poll the entire sample size (impractical as it would be anyway) -- that all you need is a smaller size, s. I ask this because I read a lot of political polls and wonder why they are always asking 1,003, 1,004, 1,005, 1,006 people the survey questions. Mike === Subject: Re: Another question for you - sample spaces answer to relates to polling and sample sizes. I seem to recall some > theorem stating that if you want to poll a sample size of S that it is > not necessary to poll the entire sample size (impractical as it would > be anyway) -- that all you need is a smaller size, s. I ask this > because I read a lot of political polls and wonder why they are always > asking 1,003, 1,004, 1,005, 1,006 people the survey questions. Suppose you choose a random sample of n people from a large population, ask a yes/no question, and receive X responses of yes. In the actual population, p is the fraction of people who would answer yes. Then X/n is the estimate of p obtained from the sample. It is a random variable with mean p and standard deviation sqrt(p (1-p)/n). Moreover, if n is fairly big and p is not too close to 0 or 1, its distribution is well approximated by the normal distribution with the same mean and standard deviation. For example, if you want the famous 3% margin of error 19 times out of 20, you might take n = 1068, which makes the standard deviation less than 0.0153, and (according to the normal distribution) the probability that |X/n - p| > 0.03 is less than 0.05. Actually, because the normal distribution is only an approximation, this is not quite right: using the true binomial distribution, that probability can be more than 0.050225, e.g. for p = 0.498165. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Another question for you - sample spaces > Hi Robert, > Something else I've been wondering about lately and can't recall the > answer to relates to polling and sample sizes. I seem to recall some > theorem stating that if you want to poll a sample size of S that it is > not necessary to poll the entire sample size (impractical as it would > be anyway) -- that all you need is a smaller size, s. I am a bit confused by your poll the entire sample size. Do you mean poll the entire population? The purpose of sampling is to get an approximate picture of an entire population without having to poll the entire population. Roughly speaking, for suitable random sampling techniques, the accuracy of your picture is roughly proportional to the square root of your sample size, but the cost of sampling is roughly proportional to the sample size itself, so the process dictates using the smallest sample that will produce the required confidence level in the results. >I ask this > because I read a lot of political polls and wonder why they are always > asking 1,003, 1,004, 1,005, 1,006 people the survey questions. Anything much less doesn't give adequate reliability and anything much more is usually too expensive. === Subject: Re: Calculator Dependence. <0001HW.C0AF42C10029730AF0284530@news.giganews.com> <448AED35.4010509@Comcast.net> Note: alt.ezines dropped, sci.math added. > I still remember those big trig and log tables in > high school/college (and beyond) and am glad to not > need them short of a power failure for my computer > or missing my solar powered scientific calculator. I've always thought it was interesting that texts continued to include these tables throughout the 1980's (and maybe even into the early 1990's), long after no one was ever using them. Well, *almost* no one. I sometimes used them to help students understand certain concepts in a more concrete fashion. For example, properties of logarithms such as log(ab) = log(a) + log(b) and log(a^b) = b*log(a) become more personalized when you spend a few minutes using log tables to perform a numerical computation such as the fifth root of 3500 times the cube root of 260. Also, the rule that 10^log(a) = a gets reinforced when you're using the antilog tables. I once gave a talk at a T^3 (Teachers Teaching with Technology) conference whose theme was a little antithetical to the party line. One part of my talk dealt with the topic What has Technology Subtracted?, and before giving a number of examples (such as logarithm tables), I said: Technology has made a number of previously taught topics obsolete. Some of these obsolete topics were useful for their reinforcement of facts that we wanted students to know very well, some of these topics were useful for their development of mathematical thinking, and some of these topics do not appear to have any significant merit beyond their necessity before present technology. I focused on these two areas: A. Topics that Reinforce Certain Facts There are several topics that once reinforced certain mathematical facts, but no longer do so in the presence of current technology. The impact of these omissions has less effect on the teaching of gifted students than on the teaching of average ability students, since the former require less drill and practice. Nonetheless, I often encounter very talented students who display some striking weaknesses. I suspect that these weaknesses are primarily caused by less drill due to the omission of these topics. I discussed Numerical calculations with logarithm tables, Calculating square roots, Arithmetic operations with rational numbers, Estimation, consistency, and exact values, and then posed the question: Do we want students to be as proficient with these skills as they once were? For those skills in which the answer is YES, what should we include in the curriculum to give our students the necessary drill? B. Topics that Develop Mathematical Thinking Skills The presence of graphing and symbolic calculators has turned some problems whose solution once involved higher order thinking skills into drill problems. I discussed Curve sketching and Algebraic and numerical computations. In the latter, I discussed the re-enforcement of algebraic facts in simplifying numerical computations, such as rewriting 5*7*11*2 as (5*2)*(7*11) and (3.8 x 10^-7)/5 being evaluated via (380 x 10^-9)/5 = (10)(38)/5 x 10^-9 = (2)(38) x 10^-9 = [(2)(30) + (2)(8)] x 10^-9 = 76 x 10^-9 = 7.6 x 10^-8). Then I posed the question: How do we compensate for the absence of topics that once required higher order thinking skills, but no longer do so? > But again, I was schooled in an era when we really > learned arithmetic and reading text-based material, > when phonetics education was universal, when being > taught axiomatic geometry and calculus courses in > high school/college were the norm, and when we were > actually taught our non dumbed-down courses by > competent teachers who weren't constantly bitching > about not being overpaid enough (and, fortunately > for those teachers, were not universally dissed). Without getting into the other things you said, I disagree with the part about calculus. In the last 10 or 20 years many, many more U.S. high school students (as a percentage of the population, even) have taken calculus than in the decades before this. My high school didn't even offer calculus (well into the early 1980's, I think), but they do now. I think you'll find the number of high school students taking AP-calculus is far higher in recent years than it was 30 or more years ago, before calculators were in general use. Now, the question of whether this is a good thing or not is another issue, one that is often brought up by college faculty who have to deal with students that took calculus in high school and then wind up not_even_placing_into_precalculus. By the way, I thought Cary Kittrell's idea (in this thread) of a calculator that forces students to determine the power-of-ten order of a computation was a great idea. I think it has significant potential (financial and educational), and she should consider approaching a calculator company about it. Dave L. Renfro === Subject: Re: Calculator Dependence. > Without getting into the other things you said, > I disagree with the part about calculus. In the > last 10 or 20 years many, many more U.S. high school > students (as a percentage of the population, even) > have taken calculus than in the decades before this. > My high school didn't even offer calculus (well into > the early 1980's, I think), but they do now. I think > you'll find the number of high school students taking > AP-calculus is far higher in recent years than it > was 30 or more years ago, before calculators were > in general use. Now, the question of whether this is > a good thing or not is another issue, one that is > often brought up by college faculty who have to > deal with students that took calculus in high school > and then wind up not_even_placing_into_precalculus. AP calculus need not be axiomatically based and usually isn't. The bloom is off the AP craze so much so that better colleges like Harvard now do not accept any AP scores lower than 5. Still, many colleges today accept AP scores of 3, 4, or 5. There's a huge difference between a 3 and a 5. An honors curriculum usually teaches more depth and retention than the mile-wide, nanometer-deep AP curricula. Those HS students taking the lower quality AP calculus (especially when not axiomatic based) do not receive the mathematical education you seem to think they do. And because of AP, they do not take the higher quality courses they would have otherwise taken. Therefore, they're worse off academically. But what the hell--getting a good education doesn't matter to many of today's crowd... Lehigh doesn't seem to have high regard for HS AP courses: Placement and Advanced Placement Credit & Calculus Readiness Test There is a big difference between calculus study at Lehigh and calculus at most high schools. A solid high-school precalculus course is necessary background for calculus at Lehigh. Students need a strong foundation in functions (forms, graphs, roots) and trigonometry to really thrive in calculus. Most students who take calculus in high school are accustomed to using a graphing calculator. Calculators are not permitted in Lehigh calculus classes. Lehigh has very high standards, and calculus students are expected to learn calculus without relying on calculators. Many students find a summer course in calculus at a local community college to be helpful in bridging from high school mathematics to Lehigh calculus. We recommend that students consider beginning Lehigh calculus a semester below their advanced placement. If you have credit for Math 21 and are eligible for Math 22, consider taking Math 21; if you have credit for Math 21 and Math 22 and are eligible for Math 23, consider taking Math 22. You will relinquish some or all of your AP credit, but experience has shown that many AP courses do not provide adequate preparation for calculus at Lehigh. BTW, Lehigh accepts AP grades 4 or 5. === Subject: Re: Calculator Dependence. <0001HW.C0AF42C10029730AF0284530@news.giganews.com> <448AED35.4010509@Comcast.net One part of my talk dealt with the topic What has > Technology Subtracted?, and before giving a number > of examples (such as logarithm tables), I said: > Technology has made a number of previously taught > topics obsolete. Some of these obsolete topics were > useful for their reinforcement of facts that we > wanted students to know very well, some of these > topics were useful for their development of mathematical > thinking, and some of these topics do not appear > to have any significant merit beyond their necessity > before present technology. > I focused on these two areas: > A. Topics that Reinforce Certain Facts > B. Topics that Develop Mathematical Thinking Skills For completeness, I should add that I also discussed C. Topics that can be left out and then I went through the same three areas in the second part of my talk, What has Technology Added?. An example of topic C (generalized to things that will probably be irrelevant several decades from now as technology becomes smarter and more user-friendly) is all the calculator-dependent graphing menu knowledge a student needs to be familiar with (trace, turn off STAT-PLOTS when graphing y = f(x), various ZOOM options, etc.). Dave L. Renfro === Subject: PRESS RELEASE: BLACK STUDENT GETS THROWN IN JAIL FOR STUDYING PRESS RELEASE: BLACK STUDENT GETS THROWN IN JAIL FOR STUDYING The following a letter I sent to the president of WESTERN WASHINGTON UNIVERSITY, Karen W. Morse I hope everyone is outraged at what happened to me. It can happen to anyone. We, as people who love The Truth - that is, use it in our work must stand against this type of aggression. I was thrown into prison for studying! After running into the same university police officer for 5 times in one school year. Wherein he waited until Memorial Day Weekend to show this very unacceptable behavior from a person trusted by The University and Students. I am now trying to secure attorneys to fight this all the way. Please take your time to investigate this matter and express your outrage to the President of WESTERN & THE PRESS, yourself. This is her email address: president@wwu.edu Program your sleep,as I show on my website, for one night and you will understand the words of Martin Luther King, Jr.: Free at last! free at last! thank God Almighty, we [ OUR MINDS ] are free at last! The Author of Ueber Alles. 6.12.2006 RE: PRESS RELEASE - Notice of Impending LEGAL ACTION Against WESTERN on RACIAL DISCRIMINATION Copies to: Attorneys/Press/Professors I am a Physicist (BS 1995), a student at WESTERN WASHINGTON UNIVERSITY and I have made an important scientific discovery - The 4th Dimension. I will not get into scientific details, but Einstein's suggestion that the 4th Dimension is TIME is not correct. What I am doing here is use the Library and other facilities, while a student to help me do work to explain my scientific discovery. I am doing my own private research while a student (INDEPENDENT LEARNING) to be able make arrangements with WESTERN and Corporations on future scientific work ( that is now under Nobel consideration) I will undertake. I plan to contact Engineering, Mathematics, Physics professors at WESTERN by September 1, 2006. I have also contacted Universities around the world to alert them to my work. This new discovery will help LAW - ENFORCEMENT, so I have taken the liberty of seeking a meeting with The FBI, CIA, and NSA. I am writing to protest the constant harassment of your University Police - I name University Police, J. Alexander [ UP# 863 ]. He has persistently harassed me on campus and I believe it is because I am a Black Man. To the effect, he gave me a Criminal Trespass, THEN had me arrested and thrown in county jail (5.27.2006) my crime was studying; I have a VALID WESTERN ID CARD , valid until September 2006 - I have paid Western Washington University, tuition; myself. This mental Grief and Distress is now unbearable. Note before: University Police, J. Alexander [ UP# 863 ], has met me at that location (where he gave me the Criminal Trespass) 2 weeks back and I explained to him that with my discovery, insomnia, and INDEPENDENT work at Western - I stay up all night (DO all nighters ) to work - I need a quiet area and the university is it. I have been doing this since September 2005! I have created a website to explain my discovery, it is called : Ueber Alles - The Struggle of Kurt Schwierige - It reads like a movie, but I use this teaching technique so people can learn complex ideas with ease. My Technical Papers are listed here: (1) New Ideas in Relativity [ Physics ] (2) PI made Rational [ Physics / Mathematics ] (3) Game Theory [ Computer Systems Engineering / Machine Science / Computer Science / Neuroscience / Physics / Mathematics / 4D-Space Theory ] (4) Machine Science [ Computer Science / Neuroscience / Computer Systems Engineering ] (5) Machine Programming [ Computer Systems Engineering / Machine Science / Computer Science / Neuroscience / Physics / Mathematics / Music / Psychology ] (6) 4D-Space Games [ Video Games played via thought : Virtual Reality Video Games Made Real... Computer Systems Engineering / Machine Science / Computer Science / Neuroscience : Play With The Future, Today. ] (7) Reflection, Light & Properties of The Invisible [ Physics & Engineering ] (8) Grid Computing, Virtual Machines & Internet Inefficiency [ Computer Science ] (9) The Unified Field Theory [ Physics ] (10) The Operating System of Organic Machines [ Computer Systems Engineering / Machine Science / Computer Science / Neuroscience / Physics / Mathematics / 4D Space Theory ] I am urgently asking for your assistance on this matter. I am at my wits end and I think I do not need to justify my scientific work to University Police, I just let them understand that I am a legal student, with legal / valid ID and I am causing no trouble to anyone. I will seek Legal Counsel on the events that happened on May 27, 2006. I have tried to go and speak with an official, but was handcuffed (FOR BEING ON SCHOOL PROPERTY), was given another citation and was escorted off campus. Now, I have been made to understand, I now have to show up in court twice, this Trespass carries a maximum FINE of 90 days in JAIL and $1000 Penalty. I NOW have a CRIMINAL RECORD as a result of the events that happened on 5.27.2006. It is a real tragedy that matters should lead to this, with such an important scientific discovery - I was planning to name Western Washington University, my school for Research and Development. How to contact me is listed on my Website. The Author of Ueber Alles. [ A very Disgusted Student ] === Subject: Re: PRESS RELEASE: BLACK STUDENT GETS THROWN IN JAIL FOR STUDYING > PRESS RELEASE: BLACK STUDENT GETS THROWN IN JAIL FOR STUDYING More interesting than the story is the mix of newsgroups to which it is posted. === Subject: Derivatives of Bessel functions? I'd like to verify something I believe is true but haven't been able to find in print or on the internet. Now, I've seen many times that J' 0 (x) {that is, the first derivative of the Bessel function of the first kind of order zero) is equal to -J 1 (x). Am I correct that: J' 0 (ax) = -aJ 1(ax), where a is a constant? === Subject: Re: Derivatives of Bessel functions? >I'd like to verify something I believe is true but haven't been able to >find in print or on the internet. Now, I've seen many times that J' 0 >(x) {that is, the first derivative of the Bessel function of the first >kind of order zero) is equal to -J 1 (x). Am I correct that: >J' 0 (ax) = -aJ 1(ax), >where a is a constant? What you meant to ask is true - what you The derivative of the function J0(ax) is not the derivative of the function J0(ax), it was the derivative of J0, evaluated at ax. The value of _that_ is -J1(ax). Anyway, find a calculus book and read up on the chain rule. ************************ David C. Ullrich === Subject: Re: Derivatives of Bessel functions? > I'd like to verify something I believe is true but > haven't been able to > find in print or on the internet. Now, I've seen > many times that J' 0 > (x) {that is, the first derivative of the Bessel > function of the first > kind of order zero) is equal to -J 1 (x). Am I > correct that: > J' 0 (ax) = -aJ 1(ax), > where a is a constant? Maybe i'm misundertsanding something, but J'0 (ax) = - J1 (ax), and if it were true it must have been -aJ1(ax) = -J1 (ax) and a = 1. So the only case when you this identity is true is a = 1. === Subject: Re: Derivatives of Bessel functions? > I'd like to verify something I believe is true but > haven't been able to find in print or on the internet. > Now, I've seen many times that J' 0 (x) {that is, the > first derivative of the Bessel function of the first > kind of order zero) is equal to -J 1 (x). Am I correct that: > J' 0 (ax) = -aJ 1(ax), > where a is a constant? I tried some obvious searches, such as and came up with quite a bit, in particular the derivative formula you asked about. For example, see Special Functions and Their Applications by N N Lebedev, Dover Publications, 1972, p. 100, eq. 5.2.9. [Use Search within this book = Derivatives of Bessel functions] Dave L. Renfro === Subject: Re: Derivatives of Bessel functions? >> I'd like to verify something I believe is true but >> haven't been able to find in print or on the internet. >> Now, I've seen many times that J' 0 (x) {that is, the >> first derivative of the Bessel function of the first >> kind of order zero) is equal to -J 1 (x). Am I correct that: >> J' 0 (ax) = -aJ 1(ax), >> where a is a constant? > I tried some obvious searches, such as > and came up with quite a bit, in particular the derivative > formula you asked about. For example, see > Special Functions and Their Applications > by N N Lebedev, Dover Publications, 1972, > p. 100, eq. 5.2.9. [Use Search within this > book = Derivatives of Bessel functions] Re-reading your post, it appears that you're asking how to get from J'0(x) = -J1(x) to J'0(ax) = -a*J1(ax). Maybe this isn't what you intended to write (because anyone studying Bessel functions should know this from a first semester calculus course), but in case this is really what you were asking, use the chain rule. Dave L. Renfro === Subject: Re: Derivatives of Bessel functions? > I'd like to verify something I believe is true but haven't been able to > find in print or on the internet. Now, I've seen many times that J' 0 > (x) {that is, the first derivative of the Bessel function of the first > kind of order zero) is equal to -J 1 (x). Am I correct that: > J' 0 (ax) = -aJ 1(ax), > where a is a constant? If, for the LHS, you meant d/dx J_0(ax), then of course; that's the chain rule. But technically, J_0'(ax) is -J_1(ax). Assuming J_0' = -J_1, which I'm pretty sure I remember, too. --Ron Bruck === Subject: Re: Derivatives of Bessel functions? > I'd like to verify something I believe is true but haven't been able to > find in print or on the internet. Now, I've seen many times that J' 0 > (x) {that is, the first derivative of the Bessel function of the first > kind of order zero) is equal to -J 1 (x). Am I correct that: > J' 0 (ax) = -aJ 1(ax), > where a is a constant? Go to http://functions.wolfram.com/ and follow your nose... I got this: I'd like to verify something I believe is true but haven't been able to > find in print or on the internet. Now, I've seen many times that J' 0 > (x) {that is, the first derivative of the Bessel function of the first > kind of order zero) is equal to -J 1 (x). Am I correct that: > J' 0 (ax) = -aJ 1(ax), > where a is a constant? I am pretty sure that this is right, but I am not 100% sure. I haven't seen Bessel functions since I got my degree in math. Dave === Subject: where biology becomes mathematics Re: defining species and rate of speciation Re: Rate of Reassortment, varies <2006Jun2.102859.15687@jarvis.cs.toronto.edu Trouble with the concept of species is that it really is confined to > sexual reproduction and not that of mitosis of microbes or viral > reproduction. Whereas the concept of species should include every > biological lifeform from the smallest of virusees to the largest plants > and animals. > There are many species concepts. Some of the definitions such as the > Biological Species Concept (cannot mate and produce viable offspring) > and the Recognition Species Concept (do not recognise each other as > potential mates) assume sexual reproduction. Others such as the > Morphological Species Concept (are different in morphology) or > Genelogical Species Concept (have common ancestry). > Note that not all species concepts apply in all biological situations. > For example the cichlid fish of the great rift lakes of East Africa are > considered to be hundreds of different species, but they are perfectly > able to breed and produce viable offspring. I'm not an expert in this > area, but even I know that any attempt to come up with a single species > concept that applies in all cases will either fail or define things as > different (or the same) species when it does not appear sensible to do > so. > Ross-c What I envision for the definition of Species, in the future, will be more like mathematics than anything known to biology of today. It will derive from Quantum Mechanics of Atom Totality theory and its modus operandi of Superdeterminism. Let me outline some ideas and concepts of how it works. We will have the entire Human Genome mapped and it will be a sequence of about 3 billion A,C,G,T and this will be a mathematical number. I am as a person unique from all other persons (fingerprint) due to my unique sequence of A,C,G,T. And as such my genome is a Unique Mathematical Number somewhere in the billions. Just for example, suppose my unique biology number is 3,012,945,678. Now another person will have his/her own unique biology number say it is 3,012,888,222. Thus every person on Earth has a unique biology number because of their A,C,G,T code sequence. Now the question is since every number is unique then a formula for the human species is obtainable because all the numbers are restricted by the Human species sequence for they cannot vary larger than a parameter. Say this parameter is a variance of 3/20,000 A,C,G,T viruses and microbes. Now there maybe alot of surprizes such as Patterns of numbers which means patterns of A,C,G,T. Just as in mathematics, there are patterns of numbers-- some are primes from 1 billion to that of 20 billion (what is the smallest living A,C,G,T string? and what is the largest living string of A,C,G,T) and some are composites. So that in mathematics, we can talk about prime numbers between 1 billion and 20 billion and compare them to composites between 1 billion and 20 billion. Patterns such as these may yield valuable insights into theoretical biology. Perhaps it is impossible to have a Prime number A,C,G,T creature due to some reason. Anyway, the Genome is turning biology into a new and exciting subfield of the subject of Mathematics because the A,C,G,T is really just a large class of numbers for which we can get formulas and equations using those numbers. So we can get a formula for what it is to be a human species as opposed to a ape species or a elm tree species. And more important we can begin to tie and connect biology to mathematics and finally Quantum Mechanics. Every subject on Earth ultimately reduces to Quantum Mechanics. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, where a & b are coprime. === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square > Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, > where a & b are coprime. If (a,b) is a solution, so are (a,-b) and (b,a), so solutions come in sets of 8 (except when a b = 0). Here is one representative of each octet: (1, 2) (4, 15) (161, 442) (22920, 50369) (2706401, 21771082) (19401465356, 37495194255) (36867352480319, 241014273471602) (2713767869349580560, 6160584655684657921) (230411421854480432456639, 599328309772220125192082) (54108065494048984622972655124, 224603401396311977137287538575) (168116478779528111798480001300551839, 331803229851166469627898043962250282) (54690892794338807630186727222541308756840, 1834835894556507914114598176948580439476929) (19954369787472117896367095755263054539599807428319, 40576759762394305384329370667159957695834264439002) etc. (The list is infinite, and yes, the visual pattern that you see about the length of the digit strings does indeed continue forever.) Keyword: Elliptic Curves. (Rank 1, torsion Z/2Z x Z/2Z ) dave === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square a^4 - 4 a^2 b^2 + b^4 = c^2 <-- [notation tweaked JR] > where a & b are coprime. > If (a,b) is a solution, so are (a,-b) and (b,a), so solutions come > in sets of 8 (except when a b = 0). Here is one representative > of each octet: > (1, 2) > (4, 15) > (161, 442) > (22920, 50369) > (2706401, 21771082) > (19401465356, 37495194255) > (36867352480319, 241014273471602) > [..] and it's easy to prove by descent that every representative solution with a odd (so that b is even) is obtained from the following recurrence with a, b, c = 1, 2, 1 a' = a^4 - b^4 b' = 2.a.b.c c' = (a^2 - b^2)^2.c^2 - 2.a^2.b^2.(a^2 + b^2)^2 (Including all the subscripts makes the equations rather busy, but think of these as a_ = a_r^4 - b_r^4 etc.) To see this, observe that the original is equivalent to the following with x, y = a^2 - b^2, a.b : x^2 - 2.y^2 = z^2 x^2 + 4.y^2 = t^2 and the first implies that for some coprime integers d, e with d odd: +/- x, y = d^2 + 2.e^2, 2.d.e If we assume that a is odd then x == 4.Z + 1 and the first mod 4 requires the +ve option. Then applying the Lucas lemma to a.b = 2.c.d with (a, b) = 1 (see past posts - I can't be bothered to quote it yet again) implies that for some set of integers p, q, r, s with (p, s) = (q, r) = 1 : a, b, c, d = p.q, 2.r.s, p.r, q.s Plugging these into a^2 - b^2 = d^2 + 2.e^2 then gives: (p^2 - 2.s^2).q^2 = (p^2 + 4.s^2).r^2 so that since (q, r) = 1 we must have for some integer t : p^2 - 2.s^2, p^2 + 4.s^2 = r^2.t, q^2.t The second implies t > 0, and taking linear combinations implies t divides 3.p^2, 3.s^2 forcing t = 1 or 3. But t = 3 dividing p^2 + 4.s^2, a sum of two squares, implies t divides p, s contrary to (p, s) = 1. Thus t = 1 and we end up with the following, which is of the same form as the original : p^2 - 2.s^2 = r^2 p^2 + 4.s^2 = q^2 John R Ramsden P.S. It's very rarely possible to settle homogenous quartics equal to a square by elementary methods as above - Dave Rusin's approach directly via elliptic curves and ranks etc is much more widely applicable. === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square > Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, > where a & b are coprime. As written, the problem is totally trivial. Has it been mis-stated? Just factor the quartic. === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square Pubkeybreaker a .8ecrit : >> Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, >> where a & b are coprime. > As written, the problem is totally trivial. Has it been > mis-stated? Just factor the quartic. It is (a^2-sqrt(2)b^2)(a^2+sqrt(2)b^2). Now what? === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square > Pubkeybreaker a .8ecrit : >> Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, >> where a & b are coprime. > As written, the problem is totally trivial. Has it been > mis-stated? Just factor the quartic. > It is (a^2-sqrt(2)b^2)(a^2+sqrt(2)b^2). Now what? If I get it right, this is: (a^2-sqrt(2)b^2)(a^2+sqrt(2)b^2) = a^4 + a^2(sqrt(2)b^2) - a^2(sqrt(2)b^2) - 2b^4 = a^4 -2b^4 Now you could write: (a^2-sqrt(2)b^2)(a^2+sqrt(2)b^2) - 4 a^2 b^2 + 3b^4 = square How does this knowledge help? ---------- So far I got this (but I don't see much usefulness in it either): > Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, > where a & b are coprime. Let square = c^2. You want a subset of the integer solutions a,b,c to: a^4 - 4 a^2 b^2 + b^4 = c^2 a^2(a+2b)(a-2b) = (c+b^2)(c-b^2) -> The problem is equivalent to: Find a,b coprime so that the term (a^2+2ab)(a^2-2ab) = a^2(a+2b)(a-2b) can be written as product of two factors which have a difference of 2b^2. The problem somehow is about factorizing. Not of the sort that I would call trivial, but maybe I'm just stupid and neither get the point nor see the irony. [ If a and b were not to be coprime, you could choose b an even number and a=b/2=sqrt(c) ... ] Ulrich === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square > Pubkeybreaker a .8ecrit : >> Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, >> where a & b are coprime. As written, the problem is totally trivial. Has it been > mis-stated? Just factor the quartic. > It is (a^2-sqrt(2)b^2)(a^2+sqrt(2)b^2). Now what? I don't think so. Try (a^2 -(2-sqrt(3))b^2) ( a^2 - (2+sqrt(3))b^2) or (a^2 +sqrt(6) a b + b^2) (a^2 -sqrt(6) a b + b^2) as factorizations into quadratics. === Subject: Re: Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square <448dba28$0$7761$7a628cd7@news.club-internet.fr> > Find all solutions of a^4 - 4 a^2 b^2 + b^4 = square, >> where a & b are coprime. > > As written, the problem is totally trivial. Has it been > mis-stated? Just factor the quartic. > It is (a^2-sqrt(2)b^2)(a^2+sqrt(2)b^2). Now what? > I don't think so. > Try > (a^2 -(2-sqrt(3))b^2) ( a^2 - (2+sqrt(3))b^2) > or > (a^2 +sqrt(6) a b + b^2) (a^2 -sqrt(6) a b + b^2) > as factorizations into quadratics. Some particular solutions of this equation: a = 1, b = 2 -> a^4 - 4 a^2 b^2 + b^4 = 1^2 a = 4, b = 15 -> a^4 - 4 a^2 b^2 + b^4 = 191^2 a = 161, b = 442 -> a^4 - 4 a^2 b^2 + b^4 = 136319^2 === Subject: Realization of homomorphisms of measure algebras Hi all, Let X and Y be sigma-finite measure spaces. Denote by E_X and E_Y the corresponding sigma-algebras and M_X and M_Y the corresponding measure algebras (the quotient of the sigma-algebra of sets by the ideal of null measure sets). Let p:M_Y -> M_X be a homomorphism of Boolean algebras respecting all sups (but not necessarily measure preserving). Under what conditions is there a function X->Y realizing p, that is a function such that f^-1:E_Y->E_X descends to p when quotiented out It is clear that this is related to the Von-Neumann-Maharam lifting theorem (finding sections of the natural quotient E_X->M_X) but I have not found precise statements of the theorem. Any help? G. Rodrigues === Subject: OMEGA of the thermal field, T=2.7 K is 2/3 H^2.3/(8.Pi.G) = q_fysik H^2/(4.Pi.G) = qf = 2/3.q_fysik Hence: OMEGA_(T=2,7 K)=2/3 TIME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: exam results I recently got the results back from my US Navy advancement exam. > I scored in the 92nd > percentile, correctly answered 56.5% of the questions (200 question > exam), and scored 64.28 points on a point scale that ranges from 20 - > 80. > My (completely frivolous) question is: how many questions did the > guy who scored 80 get right? :-D > Insufficient information. However, we can conclude a few things. The > point scale (20-80) suggests that the Navy is using a scheme similar to > that of the College Board, which uses 200-800. It evidently indicates > that the Navy scores are intended to have a mean of 50 and a standard > deviation of 10. > To check this hypothesis, let's compute what percentage of the scores > should lie below 64.28 on this scale. Mathematica says: > In[1]:= < In[2]:= CDF[NormalDistribution[50,10],64.28] > Out[2]= 0.923354 > which agrees nicely with your claim that this score is in the 92nd > percentile. We can also say that a score of 80 is 3 standard deviations > above the mean: > In[3]:= CDF[NormalDistribution[50,10],80.0] > Out[3]= 0.99865 > which is around the 99.8th percentile. But we have only a single data > point concerning the actual distribution of questions answered correctly > (56.5% right is in 92nd percentile) and this is insufficient to determine > either the mean or the standard deviation, even if we assume the raw > scores are normally distributed. Hi Dave, all, Hey, I got another data point from one of my shipmates: 36.22 (on a scale from 20 - 80) 70 questions correct (out of 200) 8th percentile Does that work? Allen === Subject: Re: does a mean vector exist? Hi Mike, I agree. But I would like to put it in more simple terms. To find the mean requires two operations. 1 is to add and the other to divide. Vectors chart the continuum. There is nothing finite or discrete about them. Therefore, an infinite continuum cannot be added or divided, esp. if it is 'continuous.' There is no such thing as omega + 1 unless you generalize the omega into an arbitrary bounded structure with + 1 as another separate and discrete number. Now they are two discrete sets coarsely united with the '+' sign. There is also no such thing as ü or 1/3 of infinity either. That would not make sense. Vectors insofar as they describe direction and magnitude do not denote anything **internally discrete.*** However, what is external is a different matter. They chart the continuum. Therefore, there might not be anything call an internal 'mean vector.' There might be a mean vector if we externalize and generalize the continuum. What is internal to the vector set is inexhaustible and ever-flowing. But what is external to the vector set might be finite and discrete. If you have something that is finite and discrete externally, then there is such a thing as a mean vector. However, if what is external i.e. V = (v1 + v2 + ... + vn) / n moves into the infinite, then there is no mean vector. You would have to encase or generalize that to an external boundary set in order to do any normal finite operation, such as looking for the arithmetic mean. B.T. > Jack: If there are finitely many vectors, say v1, v2, ..., vn, then how > about > V = (v1 + v2 + ... + vn) / n ? > This could apply to any kinds of vectors, not just physical vectors in > 3-space. > If there are infinitely many vectors, extending to an integral to get a > mean value ought to be analogous to the extension of a finite sum to an > integral. But there is a crucial difference: We usually consider and > define mean value of a single function over an interval! > Without some notion of an interval, I am not clear as to how to > extend... > Best, Mike > If I take a whole bunch of vectors and integrate them, is it possible > to > find a mean vector that approximates the flow of such vectors? > e.g. > XXX > XX > XX > I will get > X > X > X Hope you understand > Jack === Subject: Re: does a mean vector exist?- Please try the keywords pattern recognition, optical character recognition, OCR, feature extraction, skeletonization, method of moments. I bet you will get dozens or even hundreds of relevant hits at Google, or at Amazon or at university library sites if you are out for literature. Vector techniques are indispensable in these arenas. But beware: 2D vectors representing points in the (real or virtual) geometric plane are different from the 2N-dimensional vectors used to represent sets of N points in the plane. These 2N-dimensional spaces are commonly denoted as configuration spaces. Skeletonizing your picture A (see your post as quoted below) so as to obtain your picture B is an operation in such abstract configuration spaces. Please be convinced that calculating weighted (*) or plain unweighted (**) arithmetic means of sets of vectors is a most often recurring operation in pattern recognition and OCR. Arithmetic means of sets of vectors in 2D just represent =mass centres= of these sets. And that is not what you intended to find, I guess. Happy studies: Johan E. Mebius (*) Weighted arithmetic mean of vectors a1...aN with weights w1...wN = Sum (wi.ai) / Sum (wi), i = 1...N. (**) Plain arithmetic mean of vectors a1...aN = Sum (wi.ai) / N, i = 1...N. >Jack .9b.8d.93.b9.81F >>Harry .9b.8d.93.b9.81F > >>If I take a whole bunch of vectors and integrate them, is it possible >>to >>find a mean vector that approximates the flow of such vectors? >>e.g. picture A >> XXX >> XX >>XX >> >> >>I will get >> picture B >> X >> X >> X >> >> >>Hope you understand >>Jack >> >yes, check out poynting vector (?) it does that. >>It sounds really good. But as I'm not dealing with EM, I am basically >>using it for 2D textual recognition. How do you represent it using >>Numerical Methods? >>Jack >Just couldn't find any references on any books and websites (googling) CC: Jack === Subject: Re: does a mean vector exist?- Correction (**) Plain arithmetic mean of vectors a1...aN = Sum (ai) / N, i = 1...N, without the weight factors wi, of course. > Please try the keywords pattern recognition, optical character > recognition, OCR, feature extraction, skeletonization, method > of moments. > I bet you will get dozens or even hundreds of relevant hits at Google, > or at Amazon or at university library sites if you are out for > literature. > Vector techniques are indispensable in these arenas. But beware: 2D > vectors representing points in the (real or virtual) geometric plane > are different from the 2N-dimensional vectors used to represent sets > of N points in the plane. > These 2N-dimensional spaces are commonly denoted as configuration spaces. > Skeletonizing your picture A (see your post as quoted below) so as to > obtain your picture B is an operation in such abstract configuration > spaces. Please be convinced that calculating weighted (*) or plain > unweighted (**) arithmetic means of sets of vectors is a most often > recurring operation in pattern recognition and OCR. > Arithmetic means of sets of vectors in 2D just represent =mass > centres= of these sets. And that is not what you intended to find, I > guess. > Happy studies: Johan E. Mebius > (*) Weighted arithmetic mean of vectors a1...aN with weights w1...wN = > Sum (wi.ai) / Sum (wi), i = 1...N. > (**) Plain arithmetic mean of vectors a1...aN = Sum (wi.ai) / N, i = > 1...N. >Jack .9b.8d.93.b9.81F >>Harry .9b.8d.93.b9.81F >If I take a whole bunch of vectors and integrate them, is it possible >>to >>find a mean vector that approximates the flow of such vectors? >>e.g. > picture A >> XXX >> XX >>XX >>I will get > picture B >> X >> X >> X >>Hope you understand >>Jack >yes, check out poynting vector (?) it does that. >>It sounds really good. But as I'm not dealing with EM, I am basically >>using it for 2D textual recognition. How do you represent it using >>Numerical Methods? >>Jack >Just couldn't find any references on any books and websites (googling) === Subject: Re: does a mean vector exist? Take a right triangle. Draw the first number length. Draw the second number length in the right angle to it and find the hypothenuse. Draw the third number length in the right triangle to this hypothenuse. Continue always in right triangles to the last hypothenuse. In all steps, there exists a right triangle ower the hypothenuse which sides are the intermediate mean and the standard deviation. This is the Hilbert space! We see on the on the final drawing that the individual vectors were not all mutually orthogonal. The technique can be used even for nonorthogonal vectors. === Subject: Re: does a mean vector exist? Vectors are only mean when you treat them badly. === Subject: CRITICAL TORUS !!! In observing math , ive noticed , and im not alone, that many (proof)problems in math (especially the harder ones ) can be transformed into some kind of critical line problem. The most famous example is of course Riemann's Hypothese. But also the tanc conjecture can be rewritten as a critical line problem. Representions theorems can be rewritten as a critical line problem. For all clarity , i consider critical line problems as follows 1) a subgroup of zeros all lie on the critical line 2) none of the zeros of a subgroep lie on the critical line 3) no 3 zeros lie on a line (probably reducable to 1) or 2) so less important i guess ) Every upperbound or lowerbound conjecture is equivalent to some critical line problem You can just literally bend functions to get a critical line , if ya already have a simple critical function/figure/path SO if ya have a critical monotonic increasing polynoom like x^a , you can easily transform it to a line , bye transforming ( bending ) the function And don't forget that e.g. circles are the product of 2 complex lines. So critical problems in 2D are often easily transformable into critical line problems in 2D So the 2D equivalent problems are mainly and often simply reducable to critical lines A bit like belonging to an NP class to make a comparison.( but for proofs instead of computations ) However !!! some problems might not be expressible in 2D criticals , but in 3D criticals with the concept 3D being most general ! i mentioned before transforming polynoom to a line but in 3D we have topology joining the party; we could have a critical torus or polysigned critical torus or even more complicated stuff , not easily , or not at all , transformable to a line , or even 2D disk we all know topology is strongly related to number theory and metamathematics is getting popular i hope ive made clear why i believe so strongly in this critical theory importance. consider it a bit like NP=P , extended Riemann Hypothese and metamathematics. so i am deeply intrested in the critical torus since it is relating so much math concepts. and it relates them in only 2 words, nice :-) i do not claim to be a genius or anything , but as far as i know , this has never been studied. and as an intuitionist and constructivist , this is very appealing to me. i could continue with more than 3 dimensions , but i think that we should start with the simplest extension ( 2D --> 3D ; holes 0 --> +1 ) so does anybody know functions that have or might have a critical torus ? also this appears to me as a nice puzzle would be nice to see this online , e.g. at websites and forums this mail is not complete in the sence that the critical torus can be considered in different ways; do the zeros lie on the surface of the torus , or in the torus ? the most logical and intrestresting is according to me the critical surface of a torus , since other wise the condition is very loose and its more of a generalization to a strip , not a line. there is also probably a connection with eigenvalues and cellular automata , but thats confusing me at the moment. hope you think about it. i find it a logical question anyway.. greetz tommy1729 ps : id rather not have james harris replying :-) === Subject: Re: CRITICAL TORUS !!! > In observing math , ive noticed , and im not alone, that many (proof)problems > in math (especially the harder ones ) can be transformed into some kind of > critical line problem. > [..] You mention circles later on, and sometimes these critical line problems relate to circles or regions of convergence, i.e. a function expressed by, say, a power series converges inside and possibly on the boundary but not outside (although analytic functions expressed by continued fractions can have fractal regions of convergence). With the Riemann zeta function the critical line relates not to convergence but a kind of mirror symmetry. There's actually an identity which relates zeta(z) to zeta(1 - z), although that of itself doesn't imply that roots in the critical strip must be on the critical line - Pending a proof of the Riemann hypothesis, there's nothing to rule out roots occurring in pairs either side of the critical line. BTW the Riemann Mapping theorem states that any simply connected region other than the entire complex plane can be conformally mapped to any other region, such as a circle. But unless the mapping preserves some relevant aspect or property of the original function, it's hard to see what use it would be to line up roots on, say, a circle. Well hopefully the above waffle prompts a few further replies ;-P === Subject: Re: CRITICAL TORUS !!! ... >And don't forget that e.g. circles are the product of 2 complex lines. What??? Lee Rudolph === Subject: Re: CRITICAL TORUS !!! cmon i do deserve more attention then james factoring i hope ??! === Subject: diagonal & side-bisectors in parallelogram ABCD is a parallelogram. P is the middle of BC, Q the middle of DC. X is the bisection point of AQ & DB, Y is the bisection point of AP & DB. Show that X & Y divide DB into three identical sections. === Subject: Re: diagonal & side-bisectors in parallelogram > ABCD is a parallelogram. > P is the middle of BC, Q the middle of DC. > X is the bisection point of AQ & DB, Y is the bisection point of AP > DB. > Show that X & Y divide DB into three identical sections. I assume that you mean intersection rather than bisection. Hint: Note that the triangles BPY and DAY are similar. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Matrix form for polysign product Timothy might be happy I mentioned his polysigned numbers in another topic ? :-) further to answer a physics question where did the antimatter go , when the universe was created ?? simple the big bang is simply an antimatter-black-hole and bye absorbing antimatter , the matter got repelled ,and still is ...Hubbles constant what do you think about my critical torus Anthony ? post there if ya want greetz tommy === Subject: Re: Matrix form for polysign product > I'm unburying this thread in the hopes that you will respond. > The form: > s30 = s10 s20 + s11 s22 + s12 s21 > s31 = s11 s20 + s10 s21 + s12 s22 > s32 = s12 s20 + s10 s22 + s11 s21 . > is extensible to any dimension and provides a commutative product > definition. If we define addition and the scalar product in the obvious way, this defines a commutative, associative, and distributive real algebra. So you know it can only be either the direct sum of three real factors, or a real and a complex. It turns out to be the latter. If you take w to be a root of unity, then the basis elements are {1,w,w^2} and the elements of the algebra can be equated with a + bw + cw^2. The a part, from the 1, is the real factor and the {w,w^2} the complex factor in R+C. === Subject: Re: Matrix form for polysign product Avoid typing the same text again and again Stop wasting your time on mouse movements Open favorite web pages with a single hotkey press Record keystrokes and play them back with a single hotkey press ------------------------------ http://www30.webSamba.com/SmartStudio ------------------------------ EnergyKey Save yourself from repetitive tasks === Subject: Re: GPS equation WORKs.!! $$ How & Why GPS works. $$ [GPS REset & PREset equation (CORRECTED 12/06/06)]. $$ [Schwartzchild term corrected ..see *no typo* line below]. $$ (GPS daily REset FACTOR) = (GPS *PREset* FACTOR - VLO REset): $$ $$ = ----- = ------ = ------ = |[-------------------] - (------)| $$ $$ | R*g1 rVLO*g2 rVLO*g2 | $$ = |[ (----------) + (-----------) ] - (-------)| $$ | 2*c^2 2*c^2 2*c^2 | $$ $$ = |[-------------------------------] - (---------------------)| $$ $$ | G*M1 G*M1 7910^2 | $$ = |[(------------------) + (---------------------)] - (------)| $$ | 2*(nGPS - 1)*R*c^2 2*(nVLO - 1)*rVLO*c^2 2*c^2 | $$ $$ | Gibb's energy eG_GPS Gibb's ..eG_VLO eG_VLO | $$ = |[(--------------------) + (-----------------)] - (--------)| $$ | 2*m1*c^2 2*m2*c^2 2*m2*c^2 | $$ $$ | 1.18*10^16/m^3@GPS vVLO^2 | = |[PREgps #] - (REvlo # )| $$ = |[------------------] - (------)| $$ | 2.69*10^25/m^3@VLO 2*c^2 | = |4.4*10^-10 - 3.5*10^-10| $$ $$ | Schwartzchild rS | $$ = |[(GPS PREset Number, Ngps)] - (----------------)| $$ | 4*(nVLO - 1)*R | $$ $$ | GPS ambient COUNT, nGPS Schwartzchild rS | $$ = |[(-----------------------)] - (----------------)| $$ | VLO ambient COUNT, nVLO 4*(nVLO - 1)*R | $$ $$ | G*M1 | | rS | $$ = |(------------------)| = |(--------------)| $$ | 2*(nGPS - 1)*R*c^2 | | 4*(nGPS - 1)*R | no typo, $$ $$ = GUESS iSS, GPS daily REset FACTOR, ..REgps = 8.3*10^-11. ||> [=] ..7.2 microseconds [per] 86164 seconds [/Sideral day]. ||> GPS satellites are in circular orbits at 20200 km ||> altitude, or about 26580 km from the center of earth. $$ v1^2 = G*M1/(n - 1)*R = (6.674*10^-11)*(5.98*10^24)/(26580000) $$ = (3875 m/sec)^2 -> (meter)^2/(sec)^2 . $$ $$ vVLO^2 = G*M1/(n - 1)*R = (6.674*10^-11)*(5.98*10^24)/(6378200) $$ ..@84.3 min (5058 sec) = (7910 m/sec)^2 --> (meter)^2/(sec)^2. $$ Some CONCLUSiONs. $$ About GPS delay, Pioneer anomaly and planet precession advance: $$ (How & Why GPS works): $$ $$ Note the PREset and REset are SHOWN as if having positive SiGN. $$ [Also note final STANDARD alotment of *SiGN* is as per (n -1)]. $$ $$ Any mass can be M1 with m2 in LOW orbit @ r2 and m1 @ R, or r1. $$ You can be on M1, in m1 @ R or in m2 @ r2, ..same GPS equation. $$ $$ This is MOST about a General Universal COMMUNiCATiONs Equation. $$ $$ Subsequently, it's about a GLOBAL POSiTiONiNG SYSTEMs Equation. $$ $$ [Of course, this can't fully explain the Pioneer Anomaly, which $$ aside from a Solar System radial DENSiTY GRADiENT like EARth's, $$ ALSO has MORE amiss with *PEGGED* c instead of PROPER velocity]. $$ $$ GPS GR-predictions areN'T. They're Newton's ..due to altitude. $$ $$ BUT the funniest part is that NObody EVER noticed ..until GUESS. $$ $$ Finally, note, *NONE* of this has ANYthing to do with GR-Tivity. $$ $$ Density RATiO, dGPS/dVLO = 4.4*10^-10 = The GPS *PREset* FACTOR $$ = Density RATiO, (1.18*10^16/m^3 ..dGPS)/(2.69*10^25/m^3 ..dVLO) $$ = 0.000038 sec/86164 sec = [ GPS PREset ]. $$ $$ $$ $$ Finally (about BUOYANCY) radius r NOT velocity v is SiGNiFiCANT. $$ [Note the GPS equation ABOVE shows the VELOCiTY is an OPTiONAL]. $$ $$ [For SR, GR, even Newton-Tivity, in-vacu BEGiNs ..at SEA-LEVEL]. $$ $$ [There's 24 Orders-of-Magnitude DiFFERENCE between air & water]. $$ $$ How many Orders-of-Magnitude DiFFERENCE between air & Dirac Sea? $$ $$ Note the Dirac Sea photon is delta REST mass ..being transfered. $$ ABSORBED mass=h*fL/c^2=delta REST mass m1={E - delta eK}/c^2. $$ Which actually, equals { E - (delta LaGrangian L) - eV }/c^2. $$ Or alternatively, equals {E - LaGrangian L - (delta eV)}/c^2. $$ $$ No more typos ought help. Re: How & Why the GPS equation WORKs.!! [Schwartzchild typo fixed]. Re: GPS PREset & REset equation ..CORRECTED [6:PM] ; jUNE 12, 2006. Re: How & Why GPS & Pioneer anomaly works. [Duh, RE-]edited RePOST. === Subject: Force/Torque required to coil spring around mandrel. I have seen springs coiled using an electric drill clamped in a vise and the wire is fed by hand (so the springs aren't precise) onto a mandrel being spun by the drill. My question is how can the required torque of the drill be calculated based on material/geometric properties of the wire. Does anyone out there want to give this a shot? === Subject: Goedel's Proof and Transfinity Has anyone tried to show that even if you had an infinite set of postulates, you could could propose an even greater infinity of conjectures, and thus your set of postulates is still incomplete? Just wondering...! === Subject: Re: Goedel's Proof and Transfinity days. My association with the Department is that of an alumnus. >Has anyone tried to show that even if you had an >infinite set of postulates, you could could propose >an even greater infinity of conjectures, and thus your >set of postulates is still incomplete? Goedel's proof does not require the set of axioms of to be finite. In fact, the set of axioms to which Goedel applied it directly (that of the Principia) is infinite. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Goedel's Proof and Transfinity > Has anyone tried to show that even if you had an > infinite set of postulates, you could could propose > an even greater infinity of conjectures, and thus your > set of postulates is still incomplete? > Just wondering...! In most notions of logic, there are a countable number of statements. With any notion of semantics, the true statements are countable. Simply let your postulates be your true statements, ( if you could determine what they are!) and the system would be complete. === Subject: cross/dot product question I understand the math, but not the geomery meaning, pehaps. Q (X - cross pro., * - dot pro.): (a) shaw that A X (B X C) = (A X B) X C iff (A X C) X B = 0 (b) what's the meaning of A*B = 0 or B*C = 0 Well, the first part is easy and I did it. About the second part, I understand that it means that A or C are normal to B, but I don't have anything further to say about that. I tried to see if that makes the first part trivial, but I didn't come up with anything. Is there anything more interesting to say about (b)? === Subject: Re: Funny wrong proofs > I don't think these were made by any math students; > but > 16/64 cancel the sixes and get 1/4 > 19/95 cancel the nines and get 1/5 > 26/65 cancel the sixes and get 2/5 > 49/98 cancel the nines and get 4/8 > I don't know whether there are more like those. Find p/q = n/d, where n = 10p+h and d = h10^i+q, i = floor(log10(q))+1, for h=1,...,9; i.e., where pd = qn. A quickie Java program for all 4-digit numerators and denominators gave me these: 16/64 = 1/4 19/95 = 1/5 11/110 = 1/10 13/325 = 1/25 16/640 = 1/40 19/950 = 1/50 11/1100 = 1/100 13/3250 = 1/250 16/6400 = 1/400 19/9500 = 1/500 26/65 = 2/5 22/220 = 2/20 26/650 = 2/50 27/756 = 2/56 22/2200 = 2/200 26/6500 = 2/500 27/7560 = 2/560 33/330 = 3/30 39/975 = 3/75 33/3300 = 3/300 39/9750 = 3/750 49/98 = 4/8 44/440 = 4/40 49/980 = 4/80 44/4400 = 4/400 49/9800 = 4/800 55/550 = 5/50 55/5500 = 5/500 66/660 = 6/60 66/6600 = 6/600 77/770 = 7/70 77/7700 = 7/700 79/9875 = 7/875 83/332 = 8/32 88/880 = 8/80 83/3320 = 8/320 88/8800 = 8/800 99/990 = 9/90 99/9900 = 9/900 106/6625 = 10/625 111/1110 = 11/110 133/3325 = 13/325 138/8832 = 13/832 166/664 = 16/64 166/6640 = 16/640 199/995 = 19/95 199/9950 = 19/950 217/775 = 21/75 217/7750 = 21/750 222/2220 = 22/220 249/996 = 24/96 249/9960 = 24/960 266/665 = 26/65 266/6650 = 26/650 277/7756 = 27/756 333/3330 = 33/330 399/9975 = 39/975 416/6656 = 41/656 444/4440 = 44/440 499/998 = 49/98 499/9980 = 49/980 555/5550 = 55/550 568/8875 = 56/875 666/6660 = 66/660 777/7770 = 77/770 833/3332 = 83/332 888/8880 = 88/880 999/9990 = 99/990 1249/9992 = 124/992 1666/6664 = 166/664 1999/9995 = 199/995 2177/7775 = 217/775 2499/9996 = 249/996 2666/6665 = 266/665 4999/9998 = 499/998 The ones with trailing zeroes seem to be especially boring. === Subject: Re: Funny wrong proofs > Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > How about a wrong calculation (by a non-math student)? > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. > SOLUTION: > f'(x) = 2x + 3 > f'(1) = 5 > f'(2) = 7 > (f'(1) + f'(2))/2 = (5+7)/2 = 6. If the student had offered an explanation for what he was doing, it would have been entirely correct. Since f'(x) is a linear function in x, **and we know that the average value of a linear function is equal to the average value of its endpoints**, we can proceed as above! === Subject: Re: Funny wrong proofs > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. Actually, this question (from a math test, I presume) does not test mathematical knowledge, but rather tests the conventions of what various mathematical objects are called in English. By convention, the average rate of change of function f over interval [a,b] is the wording that is used to describe (f(b)-f(a))/(b-a). But I interpreted this wording to mean the average OF THE rate of change of function f over interval [a,b], which would mean the average of the derivative of f over interval [a,b] = the average of function f' over interval [a,b] = (f'(b)-f'(a))/(b-a). In my opinion, this is not quite fair as a math question, but maybe it is fair as a question of math nomenclature. The meaning of this question is too ambiguous to be on an important test (even for people whose native language is English). -- Mark Spahn > Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > How about a wrong calculation (by a non-math student)? > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. > SOLUTION: > f'(x) = 2x + 3 > f'(1) = 5 > f'(2) = 7 > (f'(1) + f'(2))/2 = (5+7)/2 = 6. If the student had offered an explanation for what he was doing, it would have been entirely correct. Since f'(x) is a linear function in x, **and we know that the average value of a linear function is equal to the average value of its endpoints**, we can proceed as above! === Subject: Re: Funny wrong proofs Woops. That line = (f'(b)-f'(a))/(b-a). should be = (f'(b)-f'(a))/(b-a), if f' is linear. > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. Actually, this question (from a math test, I presume) does not test mathematical knowledge, but rather tests the conventions of what various mathematical objects are called in English. By convention, the average rate of change of function f over interval [a,b] is the wording that is used to describe (f(b)-f(a))/(b-a). But I interpreted this wording to mean the average OF THE rate of change of function f over interval [a,b], which would mean the average of the derivative of f over interval [a,b] = the average of function f' over interval [a,b] = (f'(b)-f'(a))/(b-a). In my opinion, this is not quite fair as a math question, but maybe it is fair as a question of math nomenclature. The meaning of this question is too ambiguous to be on an important test (even for people whose native language is English). -- Mark Spahn > Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > How about a wrong calculation (by a non-math student)? > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. > SOLUTION: > f'(x) = 2x + 3 > f'(1) = 5 > f'(2) = 7 > (f'(1) + f'(2))/2 = (5+7)/2 = 6. If the student had offered an explanation for what he was doing, it would have been entirely correct. Since f'(x) is a linear function in x, **and we know that the average value of a linear function is equal to the average value of its endpoints**, we can proceed as above! === Subject: Re: Funny wrong proofs > Woops. That line = (f'(b)-f'(a))/(b-a). should be >= (f'(b)-f'(a))/(b-a), if f' is linear. >> PROBLEM. Find the average rate of change of the function >> f(x) = x^2 + 3x + 8 over the interval [1,2]. > Actually, this question (from a math test, I presume) > does not test mathematical knowledge, but rather tests > the conventions of what various mathematical objects > are called in English. > By convention, the average rate of change of function f > over interval [a,b] is the wording that is used to describe > (f(b)-f(a))/(b-a). But I interpreted this wording to mean > the average OF THE rate of change of function f over > interval [a,b], which would mean > the average of the derivative of f over interval [a,b] >= the average of function f' over interval [a,b] >= (f'(b)-f'(a))/(b-a). Funny, that's how I interpret it as well. But, how do you compute the average value of f' over [a,b]. Well, f' is a function, and in general, the way to find the average value of a function over an interval is to integrate the function over that integral, and divide by the length of the interval. So if the function is f', we get an average value of (int_a^b f'(x) dx) / (b-a). But, by the fundamental theorem of calculus, this is identical to (f(b) - f(a)) / (b-a). > In my opinion, this is not quite fair as a math question, > but maybe it is fair as a question of math nomenclature. > The meaning of this question is too ambiguous to be > on an important test (even for people whose native > language is English). I see no ambguity whatsoever. The meaning is the same either way. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Funny wrong proofs <128mm4ufv64a8ef@corp.supernews.com> <128mmsjc135r00@corp.supernews.com> Just to beat an already dead horse, here's another silly analogy: Andrew pulls out of his driveway in Berkeley at five miles per hour, headed for Southern California. Five hours and three hundred and fifty miles later, he pulls into his cousin's driveway at three miles per hour. Wow, says the cousin, you must not have hit any traffic at all. How fast did you average, anyway? Well, says Andrew, I started at five miles per hour, and just now I finished at three miles per hour, so I figure over the whole trip I must have averaged four miles per hour. --Tracy > Woops. That line = (f'(b)-f'(a))/(b-a). should be >= (f'(b)-f'(a))/(b-a), if f' is linear. >> PROBLEM. Find the average rate of change of the function >> f(x) = x^2 + 3x + 8 over the interval [1,2]. > Actually, this question (from a math test, I presume) > does not test mathematical knowledge, but rather tests > the conventions of what various mathematical objects > are called in English. > By convention, the average rate of change of function f > over interval [a,b] is the wording that is used to describe > (f(b)-f(a))/(b-a). But I interpreted this wording to mean > the average OF THE rate of change of function f over > interval [a,b], which would mean > the average of the derivative of f over interval [a,b] >= the average of function f' over interval [a,b] >= (f'(b)-f'(a))/(b-a). > Funny, that's how I interpret it as well. But, how do you compute the > average value of f' over [a,b]. Well, f' is a function, and in general, > the way to find the average value of a function over an interval is to > integrate the function over that integral, and divide by the length of > the interval. > So if the function is f', we get an average value of (int_a^b f'(x) dx) / > (b-a). > But, by the fundamental theorem of calculus, this is identical to (f(b) - > f(a)) / (b-a). > In my opinion, this is not quite fair as a math question, > but maybe it is fair as a question of math nomenclature. > The meaning of this question is too ambiguous to be > on an important test (even for people whose native > language is English). > I see no ambguity whatsoever. The meaning is the same either way. > -- > Dave Seaman === Subject: Re: Funny wrong proofs Not enough info given to solve the problem. What is the cousin's name? Amcwill > Just to beat an already dead horse, here's another silly analogy: > Andrew pulls out of his driveway in Berkeley at five miles per hour, > headed for Southern California. Five hours and three hundred and fifty > miles later, he pulls into his cousin's driveway at three miles per > hour. Wow, says the cousin, you must not have hit any traffic at > all. How fast did you average, anyway? > Well, says Andrew, I started at five miles per hour, and just now I > finished at three miles per hour, so I figure over the whole trip I > must have averaged four miles per hour. > --Tracy >> Woops. That line = (f'(b)-f'(a))/(b-a). should be >>= (f'(b)-f'(a))/(b-a), if f' is linear. > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. >> Actually, this question (from a math test, I presume) >> does not test mathematical knowledge, but rather tests >> the conventions of what various mathematical objects >> are called in English. >> By convention, the average rate of change of function f >> over interval [a,b] is the wording that is used to describe >> (f(b)-f(a))/(b-a). But I interpreted this wording to mean >> the average OF THE rate of change of function f over >> interval [a,b], which would mean >> the average of the derivative of f over interval [a,b] >>= the average of function f' over interval [a,b] >>= (f'(b)-f'(a))/(b-a). >> Funny, that's how I interpret it as well. But, how do you compute the >> average value of f' over [a,b]. Well, f' is a function, and in general, >> the way to find the average value of a function over an interval is to >> integrate the function over that integral, and divide by the length of >> the interval. >> So if the function is f', we get an average value of (int_a^b f'(x) dx) / >> (b-a). >> But, by the fundamental theorem of calculus, this is identical to (f(b) - >> f(a)) / (b-a). >> In my opinion, this is not quite fair as a math question, >> but maybe it is fair as a question of math nomenclature. >> The meaning of this question is too ambiguous to be >> on an important test (even for people whose native >> language is English). >> I see no ambguity whatsoever. The meaning is the same either way. >> -- >> Dave Seaman === Subject: Re: Funny wrong proofs <128jcotrj73jsc1@corp.supernews.com> What's the average depth of the Grand Canyon, rim to rim? Well, the depth at the south rim is 0, and the depth at the north rim is 0, so the average depth is (0+0)/2 = 0. Actually, it's a perfectly valid calculation as long as you first observe that the second derivative of f is constant, so the rate of change itself is a linear function, and average of the endpoints always gives the correct overall average if (and only if) a function is linear. (If it were a straight line between the north and south rims, the average depth would indeed be 0). The average rate of change means the average over the entire interval-you could have a function which happened to have extremely steep rates of change at both endpoints, but which on average went downhill instead of up. You take the average over an entire interval by computing the integral and dividing by the length, but in this case you're taking the integral of a derivative and you wind up just evaluating the function itself. > I don't get the joke. I am unfamiliar with the term > average rate of change of a function over an interval, > but the answer given seems perfectly reasonable. > At x=1 the function is changing at the rate of 5, > it changes at the rate of 7 and x=2, and the value > of the rate of change, the derivative of the function, is 2x+3, > which is linear. Hence the average value of 2x+3 > over interval [1,2] is 6. What am I misunderstanding? > What am I doing wrong? > -- Mark Spahn > > Hi all: > > I would like to have examples of funny wrong proofs. Here are two > > examples. They are real, in the sense that they were made by > > math students. > How about a wrong calculation (by a non-math student)? > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. > SOLUTION: > f'(x) = 2x + 3 > f'(1) = 5 > f'(2) = 7 > (f'(1) + f'(2))/2 = (5+7)/2 = 6. > This almost turned into an American Mathematical Monthly problem, until > I found out someone else had included my would-be problem in their > PROBLEM. Characterize all differentiable functions f(x) such that > (f'(a) + f'(b))/2 = (f(b) - f(a)) / (b - a), > for all real numbers a,b. > --- Christopher Heckman > ------= NextPart 000 0015 01C68BCA.A5BEAFF0 > X-Google-AttachSize: 2311 > I don't get the joke.  I am unfamiliar with the termaverage rate of change of a function over an interval,but the answer given seems perfectly reasonable.At x=1 the function is changing at the rate of 5,it changes at the rate of 7 and x=2, and the valueof the rate of change, the derivative of the function, is 2x+3,which is linear.  Hence the average value of 2x+3over interval [1,2] is 6.  What am I misunderstanding?What am I doing wrong? -- Mark Spahn  style=PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px
Proginoskes < examples of funny wrong proofs. Here are two
> examples. They are > real, in the sense that they were made by
> math students.

How > about a wrong calculation (by a non-math student)?

PROBLEM. Find the > average rate of change of the function
f(x) = x^2 + 3x + 8  over the > interval [1,2].

SOLUTION:
f'(x) = 2x + 3
f'(1) = 5
f'(2) = > 7
(f'(1) + f'(2))/2 = (5+7)/2 = 6.

This almost turned into an > American Mathematical Monthly problem, until
I found out someone else had > all differentiable functions f(x) such that

(f'(a) + f'(b))/2 = (f(b) - > f(a)) / (b - a),

for all real numbers > a,b.

     --- Christopher > Heckman
depth at the south rim is 0, and the depth at the north rim is 0, so > the average depth is (0+0)/2 = 0. No, that's the average change in height from one side to the other. > Actually, it's a perfectly valid calculation as long as you first > observe that the second derivative of f is constant, so the rate of > change itself is a linear function, and average of the endpoints always > gives the correct overall average if (and only if) a function is > linear. (If it were a straight line between the north and south rims, > the average depth would indeed be 0). The average rate of change means > the average over the entire interval-you could have a function which > happened to have extremely steep rates of change at both endpoints, but > which on average went downhill instead of up. You take the average > over an entire interval by computing the integral and dividing by the > length, but in this case you're taking the integral of a derivative and > you wind up just evaluating the function itself. Yes, or you could solve the differential equation. > I don't get the joke. I am unfamiliar with the term > average rate of change of a function over an interval, > but the answer given seems perfectly reasonable. > At x=1 the function is changing at the rate of 5, > it changes at the rate of 7 and x=2, and the value > of the rate of change, the derivative of the function, is 2x+3, > which is linear. Hence the average value of 2x+3 > over interval [1,2] is 6. What am I misunderstanding? > What am I doing wrong? > -- Mark Spahn > > Hi all: > > I would like to have examples of funny wrong proofs. Here are two > > examples. They are real, in the sense that they were made by > > math students. > How about a wrong calculation (by a non-math student)? > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. > SOLUTION: > f'(x) = 2x + 3 > f'(1) = 5 > f'(2) = 7 > (f'(1) + f'(2))/2 = (5+7)/2 = 6. > This almost turned into an American Mathematical Monthly problem, until > I found out someone else had included my would-be problem in their > PROBLEM. Characterize all differentiable functions f(x) such that > (f'(a) + f'(b))/2 = (f(b) - f(a)) / (b - a), > for all real numbers a,b. > ------= NextPart 000 0015 01C68BCA.A5BEAFF0 > X-Google-AttachSize: 2311 > I don't get the joke.  I am unfamiliar with the term
average rate of change of a function over an interval,but the answer given seems perfectly reasonable.At x=1 the function is changing at the rate of 5,it changes at the rate of 7 and x=2, and the valueof the rate of change, the derivative of the function, is 2x+3,which is linear.  Hence the average value of 2x+3over interval [1,2] is 6.  What am I misunderstanding?What am I doing wrong? -- Mark Spahn  style=PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px
Proginoskes < examples of funny wrong proofs. Here are two
> examples. They are > real, in the sense that they were made by
> math students.

How > about a wrong calculation (by a non-math student)?

PROBLEM. Find the > average rate of change of the function
f(x) = x^2 + 3x + 8  over the > interval [1,2].

SOLUTION:
f'(x) = 2x + 3
f'(1) = 5
f'(2) = > 7
(f'(1) + f'(2))/2 = (5+7)/2 = 6.

This almost turned into an > American Mathematical Monthly problem, until
I found out someone else had > all differentiable functions f(x) such that

(f'(a) + f'(b))/2 = (f(b) - > f(a)) / (b - a),

for all real numbers > a,b.

     --- Christopher > Heckman
I would like to have examples of funny wrong proofs. > How about: > Example from class: (bonehead math for non-majors) > lim x -->8 1/(x-8) ==> oo > Exam problem: > lim x --> 5 1/(x-5) > The answer given on the exam was: > | | | > When the answer was marked wrong, the student complained > that from the example given in class, she wasn't sure > whether the answer should be > | | | > or: > | | | > Don't you mean something like > > | / Yes, however the multiple underscores didn't come thru your browser for you to see. For you to see | | | with four underscores, two below and two above. === Subject: Re: Funny wrong proofs ETAtAhUArESnBV6zTJDV7GCh6+5dDbIoA7ECFAD5GOdcQ8Al9Q2176HuS4F/ZvBj The classic case, IMHO, is to reduce 16/64 or 19/95 to lowest terms by canceling the common digit. --OL === Subject: Re: Funny wrong proofs <128jcotrj73jsc1@corp.supernews.com I don't get the joke. I am unfamiliar with the term > average rate of change of a function over an interval, The average rate of change of f(x) over [a,b] is (f(b) - f(a)) / (b - a), and this is the formula that the student was supposed to use. (It's the basis of the derivative formula.) Perhaps you learned it with another name. --- Christopher Heckman > but the answer given seems perfectly reasonable. > At x=1 the function is changing at the rate of 5, > it changes at the rate of 7 and x=2, and the value > of the rate of change, the derivative of the function, is 2x+3, > which is linear. Hence the average value of 2x+3 > over interval [1,2] is 6. What am I misunderstanding? > What am I doing wrong? > -- Mark Spahn > > Hi all: > > I would like to have examples of funny wrong proofs. Here are two > > examples. They are real, in the sense that they were made by > > math students. > How about a wrong calculation (by a non-math student)? > PROBLEM. Find the average rate of change of the function > f(x) = x^2 + 3x + 8 over the interval [1,2]. > SOLUTION: > f'(x) = 2x + 3 > f'(1) = 5 > f'(2) = 7 > (f'(1) + f'(2))/2 = (5+7)/2 = 6. > This almost turned into an American Mathematical Monthly problem, until > I found out someone else had included my would-be problem in their > PROBLEM. Characterize all differentiable functions f(x) such that > (f'(a) + f'(b))/2 = (f(b) - f(a)) / (b - a), > for all real numbers a,b. > --- Christopher Heckman > ------= NextPart 000 0015 01C68BCA.A5BEAFF0 > X-Google-AttachSize: 2311 > I don't get the joke.  I am unfamiliar with the term
average rate of change of a function over an interval,but the answer given seems perfectly reasonable.At x=1 the function is changing at the rate of 5,it changes at the rate of 7 and x=2, and the valueof the rate of change, the derivative of the function, is 2x+3,which is linear.  Hence the average value of 2x+3over interval [1,2] is 6.  What am I misunderstanding?What am I doing wrong? -- Mark Spahn  style=PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px
Proginoskes < examples of funny wrong proofs. Here are two
> examples. They are > real, in the sense that they were made by
> math students.

How > about a wrong calculation (by a non-math student)?

PROBLEM. Find the > average rate of change of the function
f(x) = x^2 + 3x + 8  over the > interval [1,2].

SOLUTION:
f'(x) = 2x + 3
f'(1) = 5
f'(2) = > 7
(f'(1) + f'(2))/2 = (5+7)/2 = 6.

This almost turned into an > American Mathematical Monthly problem, until
I found out someone else had > all differentiable functions f(x) such that

(f'(a) + f'(b))/2 = (f(b) - > f(a)) / (b - a),

for all real numbers > a,b.

     --- Christopher > Heckman
Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. How about a wrong calculation (by a non-math student)? PROBLEM. Find the average rate of change of the function f(x) = x^2 + 3x + 8 over the interval [1,2]. SOLUTION: f'(x) = 2x + 3 f'(1) = 5 f'(2) = 7 (f'(1) + f'(2))/2 = (5+7)/2 = 6. This almost turned into an American Mathematical Monthly problem, until I found out someone else had included my would-be problem in their PROBLEM. Characterize all differentiable functions f(x) such that (f'(a) + f'(b))/2 = (f(b) - f(a)) / (b - a), for all real numbers a,b. --- Christopher Heckman === Subject: Re: Funny wrong proofs > Hi all: > I would like to have examples of funny wrong proofs. > Here are two > examples. They are real, in the sense that they > were made by > math students. > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = > a - b. > 2) Proof of the Cayley-Hamilton theorem: in order to > prove that a matrix > M is a zero of its characteristic polynomial, all you > have to do is to > notice that this polynomila is det(x.Id - M) and that > it is obvious that > det(M.Id - M) = det(0) = 0. > Jose Carlos Santos Here is one I encounterd myself when I was cheking homework assignments in a elementary combinatorics course. Proof that in a graph the distance between vertices satisfies the triangle inequality. Proof (actually given by more then one student): For two vertices u,v let ||u-v|| be the graph distance between u and v. We now want to prove that ||u-w||le ||u-v||+||v-w|| which holds for the triangle inequality for ||.|| QED. Kees === Subject: Re: Funny wrong proofs >>I would like to have examples of funny wrong proofs. ... >> ... real, in the sense that they were made by >>math students. >I don't think these were made by any math students; >but >16/64 cancel the sixes and get 1/4 >19/95 cancel the nines and get 1/5 >26/65 cancel the sixes and get 2/5 >49/98 cancel the nines and get 4/8 >I don't know whether there are more like those. >Tom How about (x^2 - 1)/(x - 1)? Cancel x^2 against x, leaving x in the numerator and nothing in the denominator, and cancel - against - leaving +. Hence one gets (x + 1)/(+ 1) = x + 1. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Vertex/Intersection Polygon Game Game: For 2 players, each with a different color of pencil/pen. First, using a straight-edge, draw a large n-gon (say, n = 6) on a piece of paper. (The n-gon does not need to be regular.) Each player draws straight line segments (with a straight-edge and their color pencil/pen) within the n-gon as follows: The first player draws a line segment from any vertex of the n-gon to any other. Players thereafter on each move draw a line segment between a vertex (of the n-gon) or an intersection point (made by the crossing of two previously drawn line-segments) and another vertex/intersection. Players start each line-segment where the other player ended his/her latest segment (point A). Players end each segment at any vertex/intersection (point B) such that: *No previously drawn line-segment connects the points A and B. *Aside from the 2 previously-drawn crossing segments (in the case of an intersection) or the two edges of the n-gon (in the case of a vertex), point B does not have any other line segments already drawn to it. *The 2 previously-drawn crossing line segments (in the case of an intersection) are of two different colors (ie the lines are made by different players). The last player able to move is the winner. (If a player wrongly believes he/she cannot move, then this player still loses.) Notes: Either player may draw a line-segment to/from the vertex where the first player started their first line-segment. Alternative rule: Perhaps a player should be required that any intersection they draw their segment to be of different color. Or maybe one player should draw to same-color intersections, the other player to different-colored intersections. I wonder which rule is the most fun. I suggest that after a line-segment is drawn to an intersection/vertex, then the player doing so draws a dot at the vertex so as to signify that the vertex/intersection not be drawn to again. Leroy Quet === Subject: Is Infinity A Divisor Of n? I will admit right now that I am less interested in the actual answer to this question than I am cynically interested in beginning a thread with the potential to get a lot of responses. :) If an integer m is defined as a divisor of n if n/m = integer, then wouldn't m = infinity be a divisor of every n too, since n/infinity = 0, which is an integer? And so, if true, the number of positive integers of n, written d(n) or tau(n), would have to be increased by 1 for every n. Actually, since there are an infinite kinds of infinity and there are an infinite kinds of 0's *, then d(n) = infinity for every n. *[On a number line, the average distance between every pair adjacent rational numbers is a different zero than the average distance between every pair of adjacent real numbers.] Leroy Quet === Subject: Re: Is Infinity A Divisor Of n? >I will admit right now that I am less interested > in the actual answer to this question than I am > cynically interested in beginning a thread with > the potential to get a lot of responses. > :) > If an integer m is defined as a divisor of n > if n/m = integer, then wouldn't m = infinity > be a divisor of every n too, since n/infinity = 0, > which is an integer? > And so, if true, the number of positive integers > of n, written d(n) or tau(n), would have to be > increased by 1 for every n. > Actually, since there are an infinite kinds of > infinity and there are an infinite kinds of 0's *, > then d(n) = infinity for every n. > *[On a number line, the average distance between > every pair adjacent rational numbers is a different > zero than the average distance between every pair > of adjacent real numbers.] > Leroy Quet A divisor can't be larger than the number itself. i.e., m cannot divide n if m > n... so even if infinity was a number it cannot be the divisor of any other numbers. === Subject: Re: Is Infinity A Divisor Of n? > I will admit right now that I am less interested > in the actual answer to this question than I am > cynically interested in beginning a thread with > the potential to get a lot of responses. > :) > If an integer m is defined as a divisor of n > if n/m = integer, then wouldn't m = infinity > be a divisor of every n too, since n/infinity = 0, > which is an integer? Then define integer m as a divisor (or factor) of integer n if and only if there is an integer r such that m*r = n. There is no need to use a bad definition if there is a better one available. === Subject: Re: Is Infinity A Divisor Of n? trivial. go back to high school and read the book this time. === Subject: Re: Is Infinity A Divisor Of n? >I will admit right now that I am less interested > in the actual answer to this question than I am > cynically interested in beginning a thread with > the potential to get a lot of responses. > :) > If an integer m is defined as a divisor of n > if n/m = integer, then wouldn't m = infinity > be a divisor of every n too, since n/infinity = 0, > which is an integer? This sort of operation is defined as a limit. There is no sensible explanation I have heard for n/infinity. Rather, you will see: Lim k->+inf (n/k) = 0 > And so, if true, the number of positive integers > of n, written d(n) or tau(n), would have to be > increased by 1 for every n. > Actually, since there are an infinite kinds of > infinity and there are an infinite kinds of 0's *, > then d(n) = infinity for every n. > *[On a number line, the average distance between > every pair adjacent rational numbers is a different > zero than the average distance between every pair > of adjacent real numbers.] That sounds like a definition for an infinitesimal and not for zero. The difference between any two distinct numbers is not zero. In a case like this, it sounds like a definition for two different versions of infinitesimals. The rational case would be a rational infinitesimal and the real case would be a real infinitesimal. Or that is what I would call them. There is probably a better and more rigorous definition for what such a tiny distance might mean. > Leroy Quet === Subject: Re: Is Infinity A Divisor Of n? > I will admit right now that I am less interested > in the actual answer to this question than I am > cynically interested in beginning a thread with > the potential to get a lot of responses. > :) > If an integer m is defined as a divisor of n > if n/m = integer, then wouldn't m = infinity > be a divisor of every n too, since n/infinity = 0, > which is an integer? > And so, if true, the number of positive integers > of n, written d(n) or tau(n), would have to be > increased by 1 for every n. > Actually, since there are an infinite kinds of > infinity and there are an infinite kinds of 0's *, > then d(n) = infinity for every n. > *[On a number line, the average distance between > every pair adjacent rational numbers is a different > zero than the average distance between every pair > of adjacent real numbers.] There are no adjacent pairs of rational numbers, or real numbers. === Subject: Re: Is Infinity A Divisor Of n? > I will admit right now that I am less interested > in the actual answer to this question than I am > cynically interested in beginning a thread with > the potential to get a lot of responses. > :) > If an integer m is defined as a divisor of n > if n/m = integer, then wouldn't m = infinity > be a divisor of every n too, since n/infinity = 0, > which is an integer? No. Infinity isn't an integer, therefore it is not a divisor. > And so, if true, the number of positive integers > of n, written d(n) or tau(n), would have to be > increased by 1 for every n. > Actually, since there are an infinite kinds of > infinity and there are an infinite kinds of 0's *, > then d(n) = infinity for every n. > *[On a number line, the average distance between > every pair adjacent rational numbers is a different > zero than the average distance between every pair > of adjacent real numbers.] > Leroy Quet === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net > Torkel Franz.8en, well known for his many Usenet posts, died of skeleton > cancer at Wednesday, April 19, at the age of 56. > > Torkel Franz.8en worked as a university lecturer at the department of > Computer Science and Electrical Engineering, at Lule.8c University of > Technology, Sweden. > He taught programming courses, mostly using Java and Prolog. He earned > Guide to Its Use and Abuse, which appeared in 2005. > > Usenet posts on this and related subjects, but he did also write posts > on many other subjects. > > Torkel's too early death is a great loss for his family, colleagues, > and Usenet friends. > > Erland Gadde > Department of Mathematics > Lule.8c University of Technology > Sweden > I hope that none of his family come to sci.math. For them to see how his > obituary has degenerated into the usual flame war would be a shame. > The dead deserve only respect. > Hitler deserves respect? > C-B Now, that' OTT. Torkel was in no way comparable to Hitler. A better comparison would be to Lord Haw-Haw. === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n?= is dead > Now, that' OTT. Torkel was in no way comparable to Hitler. A better > comparison would be to Lord Haw-Haw. I think that is very unfair. I disagreed quite strongly, and at length, with Torkel on the subject of Chaitin (Algorithmic Entropy); Torkel seemed to regard Chaitin as almost a fraud. But I thought his arguments were always cogent, and thought-provoking. I thought when he was rude it was generally deserved, and nothing I've read in this thread has changed my mind on that. -- === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net Torkel, OTOH, would be delighted. He loved stirring up and > perpetuating flame wars on sci.logic, and that was virtually all that > he did here. So this thread is a fitting epitaph to him. enligtenting posts in which he shared his views on mathematics, gave answers and help with various problems, and corrected many posters' statement that were in much need of corrections. > Of course his family, colleagues, and readers, who had other > experiences of him, would have other impressions themselves. Then why would certain readers NOT have the impression that all Franzen did was stir flames? It could be that certain other readers simply note, as I just did, that Franzen made a lot of different kinds of posts, including many very helpful ones. > They saw > good sides of him that were too often hidden here. I didn't see his good sides hidden. I saw them in inspiring abundance. > It is no respect to a person's memory, though, to lie about or > misrepresent the person that he was. And one salient fact about Torkel > was that he enjoyed being an asshole. He wasn't an asshole, so he couldn't have enjoyed being one. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net Torkel, OTOH, would be delighted. He loved stirring up and > perpetuating flame wars on sci.logic, and that was virtually all that > he did here. So this thread is a fitting epitaph to him. > enligtenting posts in which he shared his views on mathematics, gave > answers and help with various problems, and corrected many posters' > statement that were in much need of corrections. Oh, thank you for telling me that I hallucinated the last 5 years. > Of course his family, colleagues, and readers, who had other > experiences of him, would have other impressions themselves. > Then why would certain readers NOT have the impression that all Franzen > did was stir flames? (Nice trick making my qualifier virtually disappear, BTW; apparently you learned something from Torkel.) > It could be that certain other readers simply > note, as I just did, that Franzen made a lot of different kinds of > posts, including many very helpful ones. Or not. It could be that some readers read more than Torkel's latter-day contributions to sci.logic. > They saw > good sides of him that were too often hidden here. > I didn't see his good sides hidden. I saw them in inspiring abundance. > It is no respect to a person's memory, though, to lie about or > misrepresent the person that he was. And one salient fact about Torkel > was that he enjoyed being an asshole. > He wasn't an asshole, so he couldn't have enjoyed being one. No one is an asshole all the time. 'Being an asshole' refers to playing a role. Whatever the reason (because he liked the curmudgeon thing, because he suffered throughout the whole 5 years from his cancer), Torkel played that role to perfection and with obvious relish. That does not invalidate his alleged contributions to mathematics and logic, and those are what he should be primarily remembered for. Attempting to portray him as the Helpy Helperton of sci.logic, OTOH, is nothing but an attempt to substitute a grotesque sham for the man Torkel actually was, the man he wanted to be. > MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net > Torkel, OTOH, would be delighted. He loved stirring up and > perpetuating flame wars on sci.logic, and that was virtually all that > he did here. So this thread is a fitting epitaph to him. > enligtenting posts in which he shared his views on mathematics, gave > answers and help with various problems, and corrected many posters' > statement that were in much need of corrections. > Oh, thank you for telling me that I hallucinated the last 5 years. I'm disputing your claim that virtually all of Franzen's postings were stirring and perpetuating flames, since it is not true that virtually all of Franzen's postings were stirring and perpetuating flames. I don't claim that the reason your make the false claiim is that you hallucinated. I don't opine as to your reasons. > Of course his family, colleagues, and readers, who had other > experiences of him, would have other impressions themselves. > Then why would certain readers NOT have the impression that all Franzen > did was stir flames? Of course I know that. My point remains that on Usenet threads he did a lot more than stir flames, as he made many substantive posts with helpful content. > (Nice trick making my qualifier virtually disappear, BTW; apparently > you learned something from Torkel.) It was not intentional. The first time I DID include your qualifier. I fully grant that you did qualify with 'virtually'. But my point is not just that there were a few instances of non-flaming posting by Franzen, but that there were so very many that it is not even true that virtually all of his postings were stirring and perpetuating flames. (As well as, now I add, I don't know your definition of 'flame', so a more specific quantification depends on that definition.) It's somewhat notable that instead of just reiterating the second time that you claimed only 'virtual', you capitalized on that small lapse of mine to impugn me with trickery. > It could be that certain other readers simply > note, as I just did, that Franzen made a lot of different kinds of > posts, including many very helpful ones. > Or not. It could be that some readers read more than Torkel's > latter-day contributions to sci.logic. I'm referring to posts made by Franzen over a number of years, going back about fifteen years to near the end. He made a lot of very useful and helpful posts, and in certain instances was generous with his help. > They saw > good sides of him that were too often hidden here. > I didn't see his good sides hidden. I saw them in inspiring abundance. > It is no respect to a person's memory, though, to lie about or > misrepresent the person that he was. And one salient fact about Torkel > was that he enjoyed being an asshole. > He wasn't an asshole, so he couldn't have enjoyed being one. > No one is an asshole all the time. I don't find him to be an asshole at all. > 'Being an asshole' refers to > playing a role. Whatever the reason (because he liked the curmudgeon > thing, because he suffered throughout the whole 5 years from his > cancer), Torkel played that role to perfection and with obvious relish. I didn't find him to be a curmudgeon. I do see in him a certain kind of character who, in many instances, was arch, critical, and who often liked leaning ever so deftly from punches so that his opponent ending up sputtering, risibly off-balanced, throwing wilder and wilder powerless roundhouses. But Franzen was not at all an asshole, and while he was one tough cookie, I don't find 'curmudgeon' to fit him either. By the way, where did you learn that he felt suffering from cancer for five years? > That does not invalidate his alleged contributions to mathematics and > logic, and those are what he should be primarily remembered for. > Attempting to portray him as the Helpy Helperton of sci.logic, OTOH, is > nothing but an attempt to substitute a grotesque sham for the man > Torkel actually was, the man he wanted to be. I did not attempt to misportray him. You are the one that reduced his postings to virtually all flaming. But his postings were not virtually all flaming. And in plenty of instances he was helpful, and it is my impression that if one sincerely wanted to learn some math, then Franzen would help, ranging from small suggestions for a reading source to actually writing out formulas for the person, and where the person just wanted to continue claiming to be right while being far from right, then sometimes the best that can be made of that situation is for Franzen just to tell the person the truth about that fact. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net> > one salient fact about Torkel > was that he enjoyed being an asshole. > He wasn't an asshole, It takes one to know one. - famous saying > so he couldn't have enjoyed being one. Yes he could. He could enjoy it even if he was so dense that he didn't realize he was one. In fact, why would he do it if he didn't enjoy it - did someone make him? C-B > MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net one salient fact about Torkel > was that he enjoyed being an asshole. > He wasn't an asshole, > It takes one to know one. - famous saying That's an idiotic children's motto. And it works against your case, not for it, anyway. > so he couldn't have enjoyed being one. > Yes he could. He could enjoy it even if he was so dense that he didn't > realize he was one. That you and some other people who were on the wrong end of his expertise think he was an asshole doesn't make him one. > In fact, why would he do it if he didn't enjoy it > - did someone make him? Make him be an asshole? He wasn't one, so your question is without grounds. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net That you and some other people who were on the wrong end of his > expertise think he was an asshole doesn't make him one. i had no intention of responding to this thread as i felt my opinion was not proper but you are a good student in a bad position torkel was an ass i didn't realise this for a long time because he was also fairly bright and was skilled and confident in much he posted about but then it became my turn he was not aware of certain sources and several developments in the realm of proof theory as well as certain quotes by godel on the subject and when i mentioned them to a poster once he turned on me he was an ass not because i was wrong but merely his own lacks yet i received one of his infamous idiot responses now i was willing to explain and tried dilligently but he was not one to admit his failings that is why he was an ass because he had credibility and would not admit when he was in error that to me is always unexcusable maybe you never saw him blunder and were like me appreciative of his contributions but do not assume your view validates his superiority over those who call him an ass he also had the quality that once you saw his assness it became the big pink mountain that had always been there you are a good student so i want to offer advice (that you can then offer back to me in big heaping servings) pick your fights this one in my opinion you cannot win -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net That you and some other people who were on the wrong end of his > expertise think he was an asshole doesn't make him one. > i had no intention of responding to this thread > as i felt my opinion was not proper > but you are a good student > in a bad position > torkel was an ass > i didn't realise this for a long time > because he was also fairly bright > and was skilled and confident in much he posted about > but then it became my turn > he was not aware of certain sources > and several developments in the realm of proof theory > as well as certain quotes by godel on the subject > and when i mentioned them to a poster once > he turned on me > he was an ass not because i was wrong > but merely his own lacks > yet i received one of his infamous idiot responses > now i was willing to explain > and tried dilligently > but he was not one to admit his failings > that is why he was an ass > because he had credibility > and would not admit when he was in error > that to me is always unexcusable > maybe you never saw him blunder > and were > like me > appreciative of his contributions If you'll refer me to the thread, then I'll look at it and make my own judgment. It is possible that Franzen, being a human and thus having foibles as we all do, erred both on a matter of mathematics and philosophy and behaved dishonorably when the error was mentioned. But I can only judge as to any specific instances that are claimed by reading the actual posts. > but do not assume your view > validates his superiority over those who call him an ass You have not read me post any claim that he was in general superior to his detractors. For all I know, many of his detractors may be extremely admirable human beings in ways aside from their assessment of Franzen. But as to the specific activity of posting, and as a generalization, I do find that Franzen was a much greater poster than the people who have come into this thread to disparage him. > he also had the quality that once you saw his assness > it became the big pink mountain that had always been there > you are a good student > so i want to offer advice > (that you can then offer back to me in big heaping servings) > pick your fights > this one > in my opinion > you cannot win I think it is an easy one to win. I admired Franzen greatly and feel affection for him even from the distance I knew him. And there are many others who share that. And I notice that, as a generalization, the people who come to praise Franzen are usually people who I have learned from in the threads and are people whose postings demonstrate that they have taken some time and effort to learn and understand mathematics rather than to glibly and egregiously mangle the subject matter. That doesn't mean that these people don't make mistakes or even in that in some cases, in the heat of battle pride they have not committed lapses of intellectual honesty. But overall, the ones who have expressed admiration of Franzen are people who can give good answers and who know this stuff at a level of competence and who show seriousness about the subject; and as some have expressed, they admired not just Franzen's knowledge but also the humor he provided when he confronted the cranks, fluffs, poseurs, and blowhards - those who show the opposite of true aspiration to knowledge of this subject. But as I said, if you show me where Franzen messed up, and I can see that he did, then I will grant you that he did, and I'll do that for as many examples as you can adduce. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net It is no respect to a person's memory, though, to lie about or > misrepresent the person that he was. And one salient fact about Torkel > was that he enjoyed being an asshole. But how does one handle such a BS-er? We can counter both his desire to upset people and his hiding the truth (his lack of understanding) by simply calmly being formal. He once (about 2 months ago) claimed that what I had just written was idiotic rambling. I merely explained in detail what each word of each sentence meant. Not only did it disprove his accusation, it also showed the respectability of what I had just written - in stark contrast to his emotional outbursts. He never tried that trick on me again. C-B > Mike. === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <446205e1$0$9259$ed2619ec@ptn-nntp-reader01.plus.net He never tried that trick on me again. Well, maybe you noticed by now that then he'd have to try it while dying of skeletel cancer. I think any lack of recent intereaction between Franzen and you might be ascribed to factors having not exactly a whole heckuva a lot to do with his putative timidity in the face of your forensic brilliance. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <4467ea44.38348487@netnews.att.net> <44689f61.40782139@netnews.att.net> <44690bde.43706005@netnews.att.net> <4469dcc4.45310072@netnews.att.net> <446b4423.51884287@netnews.att.net > Just as your persistent contributions to threads everywhere are > carefully wiped off with toilet paper. > > I see why you feel so close to Torkel - you both have foul mouths. > > Please substantiate that charge about Franzen or retract it. > > MoeBlee > Just read the quotes that so many have praised in this thread. > 1. They make off-color remarks in anger. > 2. Torkel made off-color remarks in anger. > 4. Therefore making off-color remarks is respectable and admirable. > 5. Therefore those who make off-color remarks promote Torkel. > qed > Don't you believe in Logic? Maybe you should try CBL. Have you even > read it? > C-B It's standard axiomatic theorem-proving, except that after that I referred to CBL, an improvement that allows us to express and prove incompleteness results (formalizing the relation between a program and the function that it calculates.) > Anyway, you have not substantiated that Franzen had a foul mouth. This whole thread does - and, furthermore, that his followers delight only in his foul mouth. Every quote has been that of an angry man making personal insults against a person with whom he is supposed to be engaged in intelligent discourse! How often does a person not understand something? At least occasionally. How often did Torkel say, I don't understand that - could you explain it? Instead, he attacked the person - to avoid revealing that he didn't understand them. It was simply a matter of miscommunication, and he became angry about it. It was just his personal emotional weakness - similar to that of the Conservatives and the fame-worshippers who populate this forum. C-B > MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= <4467ea44.38348487@netnews.att.net> <44689f61.40782139@netnews.att.net> <44690bde.43706005@netnews.att.net> <4469dcc4.45310072@netnews.att.net> <446b4423.51884287@netnews.att.net > 1. They make off-color remarks in anger. > 2. Torkel made off-color remarks in anger. > 4. Therefore making off-color remarks is respectable and admirable. > 5. Therefore those who make off-color remarks promote Torkel. > qed > > Don't you believe in Logic? Maybe you should try CBL. Have you even > read it? > It's standard axiomatic theorem-proving, Then thank god you're the only one who uses standard axiomatic theorem-proving. > Anyway, you have not substantiated that Franzen had a foul mouth. > This whole thread does - and, furthermore, that his followers delight > only in his foul mouth. Please provide a quote by Franzen that demonstrates what you consider to be his foul mouth. By the way, any lack of such a quote contrasts with your use of words such as '' and ''. Not that I object to the use of those words, but rather that I note the utter hypocrisy of your claiming that Franzen has a foul mouth while you don't substantiate the claim and while your own language is itself often enough vulgar. > and, furthermore, that his followers delight > only in his foul mouth. Please provide a quote by an admirer of Franzen in which such a thing was said. > Every quote has been that of an angry man > making personal insults against a person with whom he is supposed to be > engaged in intelligent discourse! So personal insults are foul mouthed? Then your definition of 'foul mouthed' is so broad as to be egregiously misleading. By 'foul mouth' one might think he used words, as you do, such as '' and ''. So if you don't wish to smear him with a false charge, then you'd be more accurate to say that he posted personal insults against certain posters rather than charging that he had a foul mouth, which more suggests using words such as '' and '' as you use them. > How often does a person not understand something? At least > occasionally. How often did Torkel say, I don't understand that - > could you explain it? In most cases he DID offer to read people's explanations. Of course, in so many instances that just led to the person hanging himself with his own words. But usually Franzen DID actively seek to ellicit explanations from his interlocutors. > Instead, he attacked the person - to avoid > revealing that he didn't understand them. No, once the person demonstrated that they didn't know what the blazes they were talking about, then Franzen noted that fact. > It was simply a matter of > miscommunication, and he became angry about it. 1. It's not just innocent miscommunication when someone pontificates or blabbers on about that which he or she does not know a blazing thing. 2. You read him as being angry in his responses. I read him as being not angry at all, but rather quite cool headed while bemused while also allowing himself the prerogative to call a blowhard a blowhard. > It was just his > personal emotional weakness We all have foibles, but I don't know about the particulars of Franzen's emotional weaknesses. His postings do not evidence the weakness you claim of him. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= Would you think that TF's style of teaching is similar to that of a >>Zen master: appearing to be short and rude, but deep inside always >>caring and trying to the awaken the sense of logical reasoning? > Oh, the guy who says he's flotaing in the air but he's really sitting > on a platform attached to a pole? > It's not important whether that guy said he was floating or sitting: > it's only important that _the other guy_ needs to be awaken from > ignorant babbling! I am not the one who claimed that Torkel could float in the air. > trying to the awaken? You call that logical? It doesn't even make > sense! > It makes sense of a typo, which is more than most of C-B's non-typo > ignorant babbling! It makes sense of a typo? It IS a typo. What typo, if not itself? It makes sense of itself? Now isn't THAT babbling? >>-- >>---------------------------------------------------- >>Time passes, there is no way we can hold it back. >>Why then do thoughts linger, long after everything >>else is gone? >> Ryokan > the future but can always change it. > What an ignorant babbling! [Sure, everyone believed that C-B > could always change the results of Canadian 6/49 lottery for > next month!] Who's to say I might not burst into the lottery room when it is about to be drawn? Protestors burst into Walter Cronkite's live broadcast of the evening news on national television once. Lotteries are only local and probably less secure. BTW I held up a poster at Howard Dean's speech after the pivotal New Hampshire presidential primary, urging congress to question the 9/11 masterminds in US custody. That certainly changed the course of history. (Look for the video on my home page.) > Incompleteness of Life > Perhaps C-B needs to awake to the sense of logical reasoning, rather > than pondering the incompleteness of life, or that sort of thing. The Journal of Symbolic Logic seems to believe in it. That was just my philosophical side - lots of great philosophers are mathematicians, such as Russell and Chaitin, that is, like what they think they are, not what they really are. C-B > ---------------------------------------------------- > Time passes, there is no way we can hold it back. > Why then do thoughts linger, long after everything > else is gone? BTW The highest level of abstraction that I (using CBL) have been able to abstract the incompleteness results is to the fact that we cannot predict the future. Someone can always ask, Will your next answer be 'No'?. CBL can generate new variations as well. The only thing that we can be sure of is what we personally witness. The notion of the past is scientific only to the extent that man existed. There is no paradox in man not having existed at some point in the past. The real question is, will there be a point in time in the future at which mankind will not exist? Which incompleteness theorems correspond to that paradox? The dinosaurs are an example of a Finitely Enumerable set, useful for proving Rosser's theorem ar a lower level of abstraction, to define finite sets, and to improve Kleene's Arithmetic Hierarchy by removing the kludges to Predicate Calculus using quantifiers and less than. A consistent system is incomplete. (Rosser 1936) 1. PR/PR Provability is representable. 2. DIS/PR Therefore disprovability is representable (since negation is recursive) 3. ~PR => DIS If the system is complete 4. DIS => ~PR and consistent, 5. ~PR=DIS Lines 3,4 Then unprovability coincides with refutability. 6. ~PR/PR Line 2, SUB Line 5 Unprovability is representable. 7. -~P/P Incompleteness Axiom 8. -~PR/PR Unprovability is not representable Line 7, Substitute P=PR 9. False Line 6,8 qed Who else has ever produced this? Cite the publication. Only CBL has. (M # P/Q is defined to be P=Q(M) and P/Q is, there is an M such that M # P/Q where PR=provable, DIS=refutable (disprovable), P=any set.) > Ryokan > ---------------------------------------------------- === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= >>Would you think that TF's style of teaching is similar to that of a >>Zen master: appearing to be short and rude, but deep inside always >>caring and trying to the awaken the sense of logical reasoning? >Oh, the guy who says he's flotaing in the air but he's really sitting >on a platform attached to a pole? >>It's not important whether that guy said he was floating or sitting: >>it's only important that _the other guy_ needs to be awaken from >>ignorant babbling! > I am not the one who claimed that Torkel could float in the air. Do you see the Sun on the face of a man in a bright full-moon night? >trying to the awaken? You call that logical? It doesn't even make >sense! >>It makes sense of a typo, which is more than most of C-B's non-typo >>ignorant babbling! > It makes sense of a typo? It IS a typo. What typo, if not itself? > It makes sense of itself? Now isn't THAT babbling? If a typo doesn't make a sense of a typo then it's a non-sense, not a typo. Hint: with one small explanation of the typed error, the typo'ed sentence would make sense. A babbling on the other hand, can not be made meaningful, unless the author is really desired to get out of the babbling mood of his mind. For example, your statement that we can always change the future is a babbling, not a typo. Proof: your bursting-into-the-lottery-room explanation (below) doesn't show you could change the results of the lottery: *after* your arrest for attempting to disrupt, they either still use the old result or use a new one as _the_ result; so your statement is a a babbling, not a typo. Of course such is a trivial matter that normally doesn't need an elaboration. >>-- >>---------------------------------------------------- >>Time passes, there is no way we can hold it back. >>Why then do thoughts linger, long after everything >>else is gone? >> Ryokan >the future but can always change it. > What an ignorant babbling! [Sure, everyone believed that C-B > could always change the results of Canadian 6/49 lottery for > next month!] > Who's to say I might not burst into the lottery room when it is about > to be drawn? Protestors burst into Walter Cronkite's live broadcast of > the evening news on national television once. Lotteries are only local > and probably less secure. Again, how does your bursting into the room change the result of *next* month lottery? [snipped, the rest of C-B's usual babbling] -- ---------------------------------------------------- Time passes, there is no way we can hold it back. Why then do thoughts linger, long after everything else is gone? Ryokan ---------------------------------------------------- === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= he talks about Godel's Theorem not requiring self-reference, then cites > a proof that refers to a certain set not being r.e. who's proof is > where the self-reference occurs. Poor Torkel makes one attempt at > being original and he flubs it. > Without even looking up the passages you have in mind Now THAT explains a lot. > there is no > contradiction between (1) saying that the theorem does not REQUIRE > using self-reference in order to prove the theorem and (2) mentioning a > PARTICULAR proof of the theorem that does use self-reference. The problem, as I plainly pointed out, is that this is his example of a proof that doesn't use self-reference and in fact it DOES use self-reference. That's how incompleteness proofs work. Maybe if you looked it up you'd see that? How can you make statements about something without even seeing it? > Would you think that TF's style of teaching is similar to that of a > Zen master: appearing to be short and rude, but deep inside always > caring and trying to the awaken the sense of logical reasoning? > Oh, the guy who says he's flotaing in the air but he's really sitting > on a platform attached to a pole? > Except that you haven't mentioned anything said by Franzen that is even > remotely analogous to a claim of floating on air. The reference to his being a Zen Master doesn't qualify? > trying to the awaken? You call that logical? It doesn't even make > sense! > So you spotted a typo. Brilliant. The only example of a new result by Torkel that anyone has come up with is his reporting on a mistake made in a theorem proof where someone else came up with the theorem, and still another person pointed out the mistake. If that gives him credit for coming up with something new, then why can't I use the same technique? Turing's 1937 paper is full of mistakes. I spotted a number of mistakes in Godel's 1931 paper when I first read it. Do I get credit for coming up with a new incompleteness theorem? My new proofs and theorems are brand new, not copying what someone else said. Where else have you ever seen: No system is both consistent and complete because then the refutable sentences coincide with the unprovable sentences, but the former is r.e. (or representable) and the latter is not. (CB-2005) Who has ever come up with such a short proof of Rosser's 1936 result? Certainly not Rosser and Turing himself weighed in with a proof that is considerably longer and much more complex, requiring the construction of two complex programs, calling one from the other, and then drawing conclusions from the result! With CBL we have a proof that uses only known results from the Theory of Computation and simple formal Propositional Calculus. CBL created the shortest proof of Rosser's 1936 result known to the Mathematical world - and is thus preferable by Occam's Razor (and its formal enhancement, Occam/C-B's Razor.) That's the kind of thing you can't say of Torkel, no matter how hard you try or how much you regret that fact. The proof is in the pudding, and the above is the pudding. (CBL produces dozens of such proofs, as I have posted here repeatedly.) C-B > MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= Prevent confusions about Godel's Theorem? Just open his last text and > he talks about Godel's Theorem not requiring self-reference, then cites > a proof that refers to a certain set not being r.e. who's proof is > where the self-reference occurs. Poor Torkel makes one attempt at > being original and he flubs it. > Without even looking up the passages you have in mind > Now THAT explains a lot. No, since I'm not addressing the particular passage, but rather your general reasoning, which includes an incorrect principle as I said: > there is no > contradiction between (1) saying that the theorem does not REQUIRE > using self-reference in order to prove the theorem and (2) mentioning a > PARTICULAR proof of the theorem that does use self-reference. > The problem, as I plainly pointed out, is that this is his example of a > proof that doesn't use self-reference and in fact it DOES use > self-reference. That's how incompleteness proofs work. Maybe if you > looked it up you'd see that? How can you make statements about > something without even seeing it? Because I am addressing the general principle. You have not shown that Franzen denies that a PARTICULAR proof that uses self-reference does not use self-reference. > Would you think that TF's style of teaching is similar to that of a > Zen master: appearing to be short and rude, but deep inside always > caring and trying to the awaken the sense of logical reasoning? > > Oh, the guy who says he's flotaing in the air but he's really sitting > on a platform attached to a pole? > Except that you haven't mentioned anything said by Franzen that is even > remotely analogous to a claim of floating on air. > The reference to his being a Zen Master doesn't qualify? Where did Franzen claim that he is a Zen master? > trying to the awaken? You call that logical? It doesn't even make > sense! > So you spotted a typo. Brilliant. > The only example of a new result by Torkel that anyone has come up > with is his reporting on a mistake made in a theorem proof where > someone else came up with the theorem, No, what was claimed is that Franzen provided a PROOF of a theorem that someone attempted to prove previously but with a mistake in that attempt. > No system is both consistent and complete because then the refutable > sentences coincide with the unprovable sentences, but the former is > r.e. (or representable) and the latter is not. (CB-2005) certain amount of arithmetic. Anyway, with that qualification, you haven't shown anything I didn't know well before 2005 as I learned from books published decades ago. > The proof is in the pudding Yes, you are a prolific pudding maker. Or at least what you produce has the color and texture of pudding, I'll grant you that. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= of a r.e. but non-recursive set is proved by diagonalization, involving > no self-reference of any sort. Diagonalization is the formalization of self-reference. In reality, self-reference is mutiple reference (which I have noted here before), where one set is both the element and the set, one program is both the program and its input, one wff is both the wff with a free variable and the wff whose Godel number is substituted for that free variable, as used by Russell, Turing and Godel. In CBL, it is formalized as s11(I,J) which says that substitution is recursive [substitute J for the 1st of 1 free variable in wff number I] and f(I,J)->f(I,I) [if f(a,b) is recursive then f(a,a) is recursive] so easily generate the Fixed Point Theorem, the Recursion Theorem, the existance of a self-outputting program, and the rest of Recursion Theory. This is the kind of thing that Torkel could never fathom - formal, rigorous arguments. He instead played with personal insults, to the delight of those who understand nothing else (to the point of devoting a lengthy thread to it.) C-B > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) > Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Torkel =?ISO-8859-1?Q?Franz=E9n_is_dead?= >>This is not so, and the existence >>of a r.e. but non-recursive set is proved by diagonalization, involving >>no self-reference of any sort. > Diagonalization is the formalization of self-reference. That's a possible stance, though slightly odd - it implies that the usual proof of Cantor's theorem involves self-reference, for example. If you do think that any kind of diagonalization is self-reference, the existence of r.e. but non-recursive set is then proved by a proof involving self-reference. But even if we take this stance, it remains true that e.g. Kripke's proof of the incompleteness theorem involves no self-reference, and also that there are, for any consistent axiomatizable theory, infinitely many Diophantine equations to which the theory can't prove the lack of solutions - which was Torkel's main point: the incompleteness theorem implies incompleteness not only for strange self-referential sentences, but also for perfectly ordinary arithmetical statements. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= rigorous arguments. Of course you need to believe that. MoeBlee === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= rigorous arguments. > Of course you need to believe that. > MoeBlee Many of my posts are full of formal proofs (generated by CBL.) When was Torkel formal? (See my formal proof of Rosser's 1936 theorem elsewhere.) C-B === Subject: =?iso-8859-1?q?Re:_Torkel_Franz=E9n_is_dead?= was Torkel formal? Well, he provided me with a formal solution I had been seeking for a problem. > (See my formal proof of Rosser's 1936 theorem > elsewhere.) I'm in awe of your work. MoeBlee === Subject: Re: Search in equations <127usp2pv4sds22@news.supernews.com> <1280rib4l195h4f@news.supernews.com> Previous discussion: > Is there a way to search for a symbol within equations? >>This sounds like a software question. What software are you >>using to render the equations? >> I use Word 2000 and the equations are set with MathType 5.2. >Doesn't your find-and-replace command find and replace? > Yes. However is does not search within equations set with > MathType. In searching Google, it appears current technology > does not support searching in equations. That's correct. Global search & replace is a feature that's been requested from time to time, and it may make it into a future version of MathType. I do have a suggestion for you about converting the equations as you have previously said you needed to do. If you'll write me privately, we can discuss it off-line. Bob Mathews bobm at dessci.com Director of Training http://www.dessci.com/free.asp?free=news FREE fully-functional 30-day evaluation of MathType 5 Design Science, Inc. -- How Science Communicates MathType, WebEQ, MathPlayer, MathFlow, Equation Editor, TeXaide === Subject: Re: differential equation > I have seen today that there always exist a solution to the following > differential equation: > -u''+u=f on [0;1], where f is continuous, with the condition > u(0)=u(1)=0. But in practice, how can one find such a solution ? Numerically, most likely, and that would involve either - *diy*, with sw from netlib. http://netlib.org/ode/twpbvp.f - finding an integrated tool. http://www.sdynamix.com/trajectory.html and none of those suggested by educators who missed that you asked about BVP rather than IVP solutions. === Subject: JSH: Listening to investors While it may be a coincidence it's also possible that the recent drop in world stock markets is a vote of no-confidence in mathematicians. Investors may simply not be so willing to continue to trust a society that has spent so much time labeling someone like me, a crackpot, when in fact, I have more than enough proven results to back up my claims. And now, because mathematicians fought my mathematical proofs, I've gone after the factoring problem. This latest result I am increasingly certain provides the path to a solution. Smart money is already moving. You can deny all you want, but then you can also watch your investments melt away. S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) works because S is the surrogate, and not the target. So the k's are defined by the factors of the surrogate. Solving everything out gives you a quartic, but I think the know-how exists to work that quartic, and find solutions. If I am right, there are probably people working on the technical details now, going after the over $300k that RSA is offering. That ideas can have such a major impact--as already TRILLIONS of dollars have moved--is something that is beyond most of you as a concept, as to you mathematics is just some game, a way to make money, or pass away the time. But real world, trillions of dollars depend on systems that depend on mathematics. Some people are willing to trust, others are not. I say, the people who know human nature, know that it makes more sense that mathematicians as a group can lie than that one person can invite so much attention worldwide, be so cogent, have such dramatic and concrete results like my prime counting function, and just be crazy, so they know to MOVE THEIR MONEY. And later when it's clear that I am right, they'll know who lied. The Progress of Humanity is not just some stupid game, some useless idea that has no meaning. It is not some nothing that has been played out here, but the future of life on this planet. It is as big as it gets. This drama is as big as it gets. So a few trillion dollars here and there, are nothing, in comparison. James Harris === Subject: Re: JSH: Listening to investors you versus the whacky mathematicians, round 1536; they just won't give up! > While it may be a coincidence it's also possible that the recent drop > in world stock markets is a vote of no-confidence in mathematicians. thus: it's so clear, seeing the grey- hatched 3x3 submagicsquares, over the white ones; I meant, there's a border of the smallest boxes, that take a number, around the four extra 3x3s meaning also just one box-width between'em. the ones in the 5x5 sudoku, aside from the columns & rows, are worms e.g., but not exclusively. I'm sure, you can find a tonnage of this stuff on the highfalutin'net. >it's the 9x9 with 3x3 submagicsquares, >with the addition of 4 *more* 3x3s, placed so that >there's a border of empty boxes all around them; >makes it sort of 4-dimensional, and also quicker to solve. --it takes some to jitterbug! http://members.tripod.com/~american_almanac http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html http://larouchepub.com/other/2006/3322_ethanol_no_science.html http://www.wlym.com/pdf/iclc/howthenation.pdf === Subject: Re: interesting differential equations problem >I'm trying to find an analytical solution for this system of >differential equations, but it seems really difficult. Could someone You seem to be assuming all systems of ODEs _have_ an analytical solution. This is false. >Based on these conditions, the following differential equations >describe the system: >dA/dt = (k_-1 * B(t)) - (k_1 * A(t)) >dB/dt = (k_1 * A(t)) - (k_-1 * B(t)) - (k_2 * B(t) * (C_max - C(t))) >dC/dt = (k_2 * B(t) * (C_max - C(t))) >I'm mainly interested in finding C(t), although ideally I would like >all three functions. Your life would be a bit easier if you would change the origin on your C axis, that is, let's ignore C itself and work instead with D(t) = C_max - C(t); obviously D'(t) = C'(t). How would you describe the solution in the absence of formulas? You know A,B,D vary in time t, so a proper plot of the solution would show a curve in A,B,D,t space. (Three equations, four unknowns, and so a 1-dimensional solution set.) You say you would settle for the projection of this curve to the C-t (or D-t) plane. How about projecting to some other subspace? First, since the equations are autonomous (no explicit mention of t ) you can easily project to the A-B-D space by simply dividing through (and using the chain rule). Writing kp for k_1/k_2 and km for k_-1/k_2 the equations are simply dA/dD = (km / D) - kp * A/(B*D) dB/dD = kp * A/(B*D) - (km / D) - 1 The sum of these two equations shows that dA/dD + dB/dD = -1, which simply reflects the fact that A + B + C stays constant in your original problem (i.e. this curve I'm talking about lies in a hyperplane A + B + C = constant). Returning to the algebra, we may then write A + B = k0 - D for some constant k0; this gives us something to replace A with in the second equation, leaving simply dB/dD = kp * (k0 - D - B)/(B*D) - (km / D) - 1 which is a single, first-order ordinary differential equation. It describes a curve in the B-D plane; if you find it easier you can use the traditional names for coordinates in a plane: x = D, y = B and we are thinking here of y as a function of x. Then the ODE is just y' = a1 / (x*y) + a2 / y + a3 / x - 1 where a1 = kp*k0, a2 = -kp, a3 = -(kp+km) are just constants. This is a very pretty ODE and I'm sure someone must have studied it at some point, but I don't think the general case has a solution which can be expressed in terms of familiar functions. Even dropping any one of the four terms on the right leaves a very simple first-order ODE which Maple is unable to solve. Indeed, Maple can't even say anything useful about a very basic case like y' = 1/y - 1/x - 1 . (For that matter, it doesn't even comment on y' = 1/y - 1/x . Who knew!) Of course some cases are easy. If a1=a2=0 then this is an easy ODE, and if a1=a3=0 it has the general solution y(x) = a2*( 1 + LambertW(C*exp(-x/a2)) ) but I doubt this conveys much understanding even in this special case. The special case a2=a3=0 is worse; an integrating factor is exp( (x+y)^2/(2*a1) )*(y/x) but when integrating we obtain an transcendental equation to solve for y(x). Ditto for e.g. a1=0, a2=a3. As these special cases perhaps illustrate, simultaneously scaling x and y allows us to reduce to the case in which, say, a2 = 1 (or 0) but no other simple transformation of the variables seems to be very useful. dave